Infinite limits
Given (s_n) divergent succession to +\infty, we will denote it as:
\underset {n \to \infty} {\lim} s_n = +\infty if \forall M \in \mathbb{R}, \exists N \in \mathbb{N} |n < \overset{n}{N}\Rightarrow s_n > M
Given (s_n) divergent succession to -\infty, we will denote it as:
\underset {n \to \infty} {\lim} s_n = -\infty if \forall M \in \mathbb{R}, \exists N \in \mathbb{N} |n > N\Rightarrow s_n < \overset{n}{M}
A divergent succession is a sequence that diverges to +\infty or -\infty
The successions that are not convergent nor divergent are called oscillating
Henceforth, we will say that a succession has limit if it is convergent or divergent, that is, it is not oscillating
The limit remains unique:
If a, b \in \mathbb{R} \cup \left \{ +\infty, -\infty \right \} \left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_n = a
\\ \underset {n \to \infty} {\lim} s_n = b
\end{matrix}\right.
\Rightarrow a = b
To the inheritance of convergent we will refer to as finite limit successions and to the divergent successions as successions with infinite limit
From the definition of divergent succession it can be inferred immediately that if a succession is monotonically non decreasing or not growing and not bounded above or inferiorly then diverges to +\infty or -\infty
Proposition
Every monotone succession has a limit (finite if bounded, infinite otherwise)
Example
We saw earlier that the succession H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots of n-th term was monotonous, strictly growing and was not bounded above
Then it is divergent and by the proposition, it will have limit, which in this case will be +\infty
Results on divergent successions
- Any sub-succession of a divergent succession to
+\infty or -\infty is divergent to +\infty or -\infty
-
A succession has a divergent sub-succession to +\infty or -\infty if and only if it is not bounded above or below
In general, a succession has a divergent sub-succession, if and only if it is not bounded
- If
(s_n) is a divergent succession to +\infty or -\infty and (t_n) is a succession bounded lowerly or higher, the succession (s_n + t_n) diverges to +\infty or -\infty
-
If (s_n) is a divergent succession to +\infty or -\infty and (t_n) is a convergent or divergent succession to +\infty or -\infty, the succession (s_n + t_n) diverges to +\infty or -\infty. The abbreviated form:
\tiny\left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_n = +\infty, \underset {n \to \infty} {\lim} t_n = a \in \mathbb{R}\cup +\infty \Rightarrow \underset {n \to \infty} {\lim} \left ( s_n + t_n \right ) = +\infty \\
\underset {n \to \infty} {\lim} s_n = -\infty, \underset {n \to \infty} {\lim} t_n = a \in \mathbb{R}\cup -\infty \Rightarrow \underset {n \to \infty} {\lim} \left ( s_n + t_n \right ) = -\infty
\end{matrix}\right.
- The sum of a succession divergent to
+\infty with a divergent succession to -\infty can be convergent, divergent to +\infty, diverging the -\infty or oscillating
- If
(s_n) is a divergent succession to +\infty or -\infty and (t_n) is a succession with r > 0 and m such that t_n > r always that n > m, the succession (s_n \cdot t_n) diverges to +\infty or -\infty
- If
(s_n) is a divergent succession to +\infty or -\infty and (t_n) is a succession with r > 0 and m such that t_n < -r always that n > m, the succession (s_n \cdot t_n) diverges to -\infty or +\infty
-
If (s_n) is a divergent succession to +\infty or -\infty and (t_n) is a convergent succession with positive or divergent limit to +\infty or -\infty, the succession (s_n \cdot t_n) diverges to +\infty. The abbreviated form:
\tiny\left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_n = +\infty, \underset {n \to \infty} {\lim} t_n = a \in \mathbb{R}\cup \left (0, +\infty \right )\cup +\infty \Rightarrow \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right ) = +\infty \\
\underset {n \to \infty} {\lim} s_n = -\infty, \underset {n \to \infty} {\lim} t_n = a \in \mathbb{R}\cup \left (0, +\infty \right )\cup -\infty \Rightarrow \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right ) = +\infty
\end{matrix}\right.
-
If (s_n) is a divergent succession to -\infty or +\infty and (t_n) is a convergent succession with positive or divergent limit to +\infty or -\infty, the succession (s_n \cdot t_n) diverges to -\infty. The abbreviated form:
\tiny\left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_n = -\infty, \underset {n \to \infty} {\lim} t_n = a \in \mathbb{R}\cup \left (0, +\infty \right )\cup -\infty \Rightarrow \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right ) = -\infty \\
\underset {n \to \infty} {\lim} s_n = +\infty, \underset {n \to \infty} {\lim} t_n = a \in \mathbb{R}\cup \left (0, +\infty \right )\cup +\infty \Rightarrow \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right ) = -\infty
\end{matrix}\right.
- The product of a divergent succession to
+\infty or -\infty by a convergent succession to 0 can be convergent, divergent to +\infty, diverging the -\infty or oscillating
-
If (s_n) diverges to +\infty or -\infty if and only if it has at most a finite number of non positive terms and its inverse converges to 0. Or in short form:
\tiny\left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_n = +\infty \left\{\begin{matrix}
\exists m | s_n > 0 \text{ siempre que }n > m\\
\underset {n \to \infty} {\lim} \frac{1}{s_n} = 0
\end{matrix}\right.\\
\underset {n \to \infty} {\lim} s_n = -\infty \left\{\begin{matrix}
\exists m | s_n < 0 \text{ siempre que }n > m\\
\underset {n \to \infty} {\lim} \frac{1}{s_n} = 0
\end{matrix}\right.
\end{matrix}\right.
-
The succession of absolute values of a succession (s_n) diverges to +\infty if and only if it has at most a finite number of non zero terms and its inverse converges to 0. Or in short form:
\tiny\underset {n \to \infty} {\lim} \left \| s_n \right \| = +\infty \Leftrightarrow \left\{\begin{matrix}
\exists m | s_n \neq 0 \text{ siempre que } n > m \\
\underset {n \to \infty} {\lim} \frac{1}{s_n} = 0
\end{matrix}\right.
-
It is easy to check that a succession (s_n) converges to 0 if and only if the succession (\left \| s_n \right \|) of its absolute values converges to 0
Both properties are equivalent to \forall \epsilon > 0, \exists N| \left \| s_n \right \| < \epsilon for n > N
In general, it can only be said that if (s_n) is convergent with limit a, then \left ( \left \| s_n \right \| \right ) is convergent with limit \left \| a \right \|; the reciprocal is not always true if a\neq 0
So we can deduce that a succession (s_n) without null terms converges to 0 if and only if the succession \frac{1}{\left \| s_n \right \|} from the inverse absolute values, it diverges to +\infty. The abbreviated form:
If s_n\neq 0, \forall n|\underset {n \to \infty} {\lim}s_n = 0\Leftrightarrow \underset {n \to \infty} {\lim}\frac{1}{\left \| s_n \right \|} =+\infty
-
Given that \frac{s_n}{t_n} = s_n \cdot \frac{1}{t_n}, the above results on the product, can be easily applied on the quotient
For example, if (s_n) is a divergent succession to +\infty and (t_n) is a positive limit succession or a convergent succession to 0 that has a finite number of non positive terms, then \left (\frac{s_n}{t_n} \right ) is a divergent succession to +\infty
-
If the quotient succession \left (\frac{s_n}{t_n} \right ) is defined, can be convergent, divergent to +\infty, diverging the -\infty or oscillating, if we are in any of these cases:
(s_n) and (t_n) converge to 0\underset {n \to \infty} {\lim} \left \| s_n \right \| = \underset {n \to \infty} {\lim} \left \| t_n \right \| = +\infty
If (s_n) has non zero limit (finite or infinite) and (t_n) converges to 0, its quotient is divergent except in the case that it has infinite positive terms and infinite negative terms
-
Fitting divergent successions
Given two successions (s_n) and (t_n) such that \exists m| s_n\leq t_n; \forall n > m it is verified that:
-
If (s_n) diverges to +\infty, (t_n) it also diverges to +\infty. The abbreviated form:
\underset {n \to \infty} {\lim} s_n = +\infty\Rightarrow \underset {n \to \infty} {\lim} t_n = +\infty
-
If (t_n) diverges to -\infty, (s_n) it also diverges to -\infty. The abbreviated form:
\underset {n \to \infty} {\lim} t_n = -\infty\Rightarrow \underset {n \to \infty} {\lim} s_n = -\infty
Extended straight
Thanks to the above results we can expand the set of reals as follows:
\mathbb{\overline{R}}=\mathbb{R}\cup {+\infty,-\infty}
To this structure \mathbb{\overline{R}} is often referred to as extended straight
Like the order structure of \mathbb{R} and (partially) its algebraic operations are extended as follows:
\forall x\in\mathbb{\overline{R}}|-\infty \leqslant x \leqslant +\infty\forall x\in\mathbb{\overline{R}}, x\neq -\infty|(+\infty) + x =\ x + (+\infty) = +\infty-
\forall x\in\mathbb{\overline{R}}, x\neq +\infty|(-\infty) + x =\ x + (-\infty) = -\infty
Remaining to be defined (+\infty) + (-\infty) and (-\infty) + (+\infty)
-(+\infty) = -\infty, -(-\infty) = +\infty-
\forall x, y\in\mathbb{\overline{R}}|x - y = x + (-y)
As long as the sum makes sense and remains to be defined (+\infty) - (+\infty) and (-\infty) - (-\infty)
\forall x\in (0, +\infty)\cup \left \{ +\infty \right \} | (+\infty) \cdot x = x \cdot (+\infty) = +\infty
\forall x\in \left \{ -\infty \right \} \cup (-\infty, 0) | (+\infty) \cdot x = x \cdot (+\infty) = -\infty
\forall x\in (0, +\infty) \cup \left \{ +\infty \right \} | (-\infty) \cdot x = x \cdot (-\infty) = -\infty
-
\forall x\in \left \{ -\infty \right \} \cup (-\infty, 0) | (-\infty) \cdot x = x \cdot (-\infty) = +\infty
Remaining to be defined (+\infty) \cdot 0, 0 \cdot (+\infty), (-\infty) \cdot 0 and 0 \cdot (-\infty)
\frac{1}{+\infty} = \frac{1}{-\infty} = 0
-
\forall x, y \in \mathbb{\overline{R}}|\frac{x}{y} = x \cdot \left (\frac{1}{y} \right )
As long as the product makes sense and remains to be defined \frac{1}{0} and \frac{x}{0}, \forall x \in \mathbb{\overline{R}}, así como \frac{+\infty}{+\infty}, \frac{+\infty}{-\infty}, \frac{-\infty}{+\infty} and \frac{-\infty}{-\infty}
\left \| +\infty \right \| = \left \| -\infty \right \| = +\infty
Properties of the extended straight
In \mathbb{\overline{R}} there are the top and bottom quotas of a non empty set, just like supreme, tiny, maximum and minimum
Every non empty subset of \mathbb{\overline{R}} has supreme (minimum top quota) and tiny (maximum lower bound) in \mathbb{\overline{R}}
-
Dice x, y \in \mathbb{\overline{R}} | x < y, we can find z \in \mathbb{R} | x < z < y
In other words, (x,y)\subseteq \mathbb{\overline{R}} contains real numbers
-
Dice x, y, z \in \mathbb{\overline{R}}|x\geq y\Rightarrow x+z\geq y+z
As long as the sums are defined
\forall x \in \mathbb{\overline{R}}|(-1)\cdot x = -x- In
\mathbb{\overline{R}} they continue to be verified all the properties of the absolute value, provided that the operations are defined
-
Given a succession (s_n) with limit a (finite or infinite) and a succession (t_n) with limit b (finite or infinite), we have to:
- If
a + b is set to \mathbb{\overline{R}}, (s_n + t_n) has limit a + b
- If
a - b is set to \mathbb{\overline{R}}, (s_n - t_n) has limit a - b
- If
a \cdot b is set to \mathbb{\overline{R}}, (s_n \cdot t_n) has limit a \cdot b
- If
\frac{a}{b} is set to \mathbb{\overline{R}}, (\frac{s_n}{t_n}) has limit \frac{a}{b}
Oscillation limits
Upper limit
It (s_n) a top bounded succession
The succession (x_n)\in \mathbb{R} defined as x_n = \text{sup }\left \{ s_k|k\geq n \right \} is monotone not increasing, so it has limit (which can be finite or -\infty)
This limit is called upper limit of the succession (s_n) and it is denoted \tiny\left\{\begin{matrix}
\underset {n \to \infty} {\lim} \text{sup } s_n = \underset {n \to \infty} {\lim} \left ( \underset{k\geq n} {\text{sup } } s_n \right ) = \underset {n \to \infty} {\text{inf } } \left ( \underset{k\geq n} {\text{sup } } s_k \right ) \\
\underset {n \to \infty} {\lim} \text{sup } s_n = + \infty && \text{si no est}{a}'\text{ acotada superiormente}
\end{matrix}\right.
Lower limit
It (s_n) a lower bounded succession
The succession (y_n)\in \mathbb{R} defined as y_n = \text{inf }\left \{ s_k|k\geq n \right \} is monotone not decreasing, so it has limit (which can be finite or +\infty)
This limit is called lower limit of the succession (s_n) and it is denoted \tiny\left\{\begin{matrix}
\underset {n \to \infty} {\lim} \text{inf } s_n = \underset {n \to \infty} {\lim} \left ( \underset{k\geq n} {\text{inf } } s_n \right ) = \underset {n \to \infty} {\text{sup } } \left ( \underset{k\geq n} {\text{inf } } s_k \right ) \\
\underset {n \to \infty} {\lim} \text{sup } s_n = - \infty && \text{si no est}{a}'\text{ acotada inferiormente}
\end{matrix}\right.
An immediate consequence of both definitions is that \underset {n \to \infty} {\lim} \left ( \underset{k\geq n} {\text{inf } } s_n \right ) \leq \underset {n \to \infty} {\lim} \left ( \underset{k\geq n} {\text{sup } } s_n \right )
Examples of lower and upper limits
In some of these examples it is not easy to rigorously demonstrate what the lower and upper bounds are
- Given the succession
(-1)^n, \underset {n \to \infty} {\lim} \text{inf } (-1)^n = -1 and \underset {n \to \infty} {\lim} \text{sup } (-1)^n = 1
- Given the succession
(-1)^n \cdot n, \underset {n \to \infty} {\lim} \text{inf } (-1)^n \cdot n = -\infty and \underset {n \to \infty} {\lim} \text{sup } (-1)^n \cdot n = \infty
- Given the succession
\frac{(-1)^n}{n}, \underset {n \to \infty} {\lim} \text{inf } \frac{(-1)^n}{n} = 0 and \underset {n \to \infty} {\lim} \text{sup } \frac{(-1)^n}{n} = 0
- Given the succession
\text{sen } n, \underset {n \to \infty} {\lim} \text{inf } \text{sen } n = -1 and \underset {n \to \infty} {\lim} \text{sup } \text{sen } n = 1
- Given the succession
(0, 1, 0, 1, 0, 1, \cdots), \underset {n \to \infty} {\lim} \text{inf } (0, 1, 0, 1, 0, 1, \cdots) = 0 and \underset {n \to \infty} {\lim} \text{sup } (0, 1, 0, 1, 0, 1, \cdots) = 1
- Given the succession
(0, 1, 0, 2, 0, 3, \cdots), \underset {n \to \infty} {\lim} \text{inf } (0, 1, 0, 2, 0, 3, \cdots) = 0 and \underset {n \to \infty} {\lim} \text{sup } (0, 1, 0, 2, 0, 3, \cdots) = +\infty
- Given the succession
(0, \frac{1}{2}, 0, \frac{1}{3}, 0, \frac{1}{4}, \cdots), \underset {n \to \infty} {\lim} \text{inf } (0, \frac{1}{2}, 0, \frac{1}{3}, 0, \frac{1}{4}, \cdots) = 0 and \underset {n \to \infty} {\lim} \text{sup } (0, \frac{1}{2}, 0, \frac{1}{3}, 0, \frac{1}{4}, \cdots) = 0
- Given the succession
(a, b, c, a, b, c, \cdots), \underset {n \to \infty} {\lim} \text{inf } (a, b, c, a, b, c, \cdots) = \left \{ a, b, c \right \} and \underset {n \to \infty} {\lim} \text{sup } (a, b, c, a, b, c, \cdots) = \text{m}\acute{a}\text{x}\left \{ a, b, c \right \}
Proposition 1
Given a succession s_n you have to:
- It is convergent with limit a if and only if
\underset {n \to \infty} {\lim} \text{inf } s_n = \underset {n \to \infty} {\lim} \text{sup } s_n = a
- It is divergent with limit
+\infty if and only if \underset {n \to \infty} {\lim} \text{inf } s_n = +\infty
- It is divergent with limit
-\infty if and only if \underset {n \to \infty} {\lim} \text{sup } s_n = \underset {n \to \infty} {\lim} \text{inf } s_n = -\infty
Demonstration of proposition 1 A)
Dice \forall n \in \mathbb{N}; x_n = \text{sup }\left \{ s_k|k\geqslant n \right \}; y_n = \text{inf }\left \{ s_k|k\geqslant n \right \}
From this definition we have to y_n \leq s_n \leq x_n
As \underset {n \to \infty} {\lim} \text{inf } s_n = \underset {n \to \infty} {\lim} y_n and \underset {n \to \infty} {\lim} \text{sup } s_n = \underset {n \to \infty} {\lim} x_n, si \underset {n \to \infty} {\lim} \text{inf } s_n = \underset {n \to \infty} {\lim} \text{sup } s_n = a \in \mathbb{R} nos basta con aplicar la regla del Sandwich para obtener que s_n es convergente con límite a
Reciprocally, if s_n is convergent with limit a, given \epsilon > 0, exists an N such that \forall n >N is true that a - \frac{\epsilon}{2} < s_n < a + \frac{\epsilon}{2}
So \forall n >N the set \left \{ s_k|k\geq n \right \} is bounded at the top for a + \frac{\epsilon}{2} e inferiormente para a - \frac{\epsilon}{2}
And we have to \forall n >N is true that a - \epsilon < a- \frac{\epsilon}{2} \leq y_n \leq s_n \leq a + \frac{\epsilon}{2} < a + \epsilon
So it is also true that \underset {n \to \infty} {\lim} \text{inf } s_n = \underset {n \to \infty} {\lim} y_n = a = \underset {n \to \infty} {\lim} x_n = \underset {n \to \infty} {\lim} \text{sup } s_n which is what we wanted to demonstrate
Demonstration of proposition 1B)
We need \underset {n \to \infty} {\lim} \text{inf } s_n = +\infty so s_n should be bounded below
Dice \forall n \in \mathbb{N}; x_n = \text{sup }\left \{ s_k|k\geqslant n \right \}; y_n = \text{inf }\left \{ s_k|k\geqslant n \right \}
From this definition we have to y_n \leq s_n
Which forces s_n is also divergent to +\infty
Reciprocally, if s_n is divergent to +\infty, then it is not bounded above and by definition it is true that \underset {n \to \infty} {\lim} \text{sup } s_n = +\infty
So \underset {n \to \infty} {\lim} \text{inf } s_n = +\infty since given M \in \mathbb{R} there exists an N such that \forall n >N is true that s_n > M + 1
So it is also true that y_n \geq y_N \geq M + 1 > M and we have to \underset {n \to \infty} {\lim} \text{inf } y_n = +\infty which is what we wanted to demonstrate
Demonstration of proposition 1 C)
We need \underset {n \to \infty} {\lim} \text{sup } s_n = -\infty so s_n debe ser acotada superiormente
Dice \forall n \in \mathbb{N}; x_n = \text{sup }\left \{ s_k|k\geqslant n \right \}; y_n = \text{inf }\left \{ s_k|k\geqslant n \right \}
From this definition we have to s_n \leq x_n
Which forces s_n is also divergent to -\infty
Reciprocally, if s_n is divergent to -\infty, then it is not bounded below and by definition it is true that \underset {n \to \infty} {\lim} \text{inf } s_n = -\infty
So \underset {n \to \infty} {\lim} \text{sup } s_n = -\infty since given M \in \mathbb{R} there exists an N such that \forall n >N is true that s_n < M + 1
So it is also true that M + 1 > M \geq x_N \geq x_n and we have to \underset {n \to \infty} {\lim} \text{sup } x_n = -\infty which is what we wanted to demonstrate
Corollary of Proposition 1
A succession (s_n) has limit on \mathbb{\overline{R}} if and only if \underset {n \to \infty} {\lim} \text{inf }s_n = \underset {n \to \infty} {\lim} \text{sup }s_n
And in this case, the limit is equal to the upper limit and the lower limit
The succession (s_n) is oscillating if and only if \underset {n \to \infty} {\lim} \text{inf }s_n < \underset {n \to \infty} {\lim} \text{sup }s_n
Demonstration of the Corollary of Proposition 1
From Proposition 1 A) we see that \underset {n \to \infty} {\lim} \text{inf }s_n = \underset {n \to \infty} {\lim} \text{sup }s_n it is immediate since \underset {n \to \infty} {\lim} \text{inf } s_n = \underset {n \to \infty} {\lim} \text{sup } s_n = a
And in case it is oscillating we need that \underset {n \to \infty} {\lim} \text{inf } s_n < \underset {n \to \infty} {\lim} \text{sup } s_n
Dice \forall n \in \mathbb{N}; x_n = \text{sup }\left \{ s_k|k\geqslant n \right \}; y_n = \text{inf }\left \{ s_k|k\geqslant n \right \}
From this definition we have to y_n \leq s_n \leq x_n
And to make it oscillating y_n \neq s_n \neq x_n so we are left with that y_n < s_n < x_n
So it is also true that y_n < x_n and we have to \underset {n \to \infty} {\lim} \text{inf } s_n < \underset {n \to \infty} {\lim} \text{sup } s_n which is what we wanted to demonstrate
Oscillation limit
A number is said to a \in \mathbb{\overline{R}} it is a oscillation limit of a succession s_n if a is the limit of some sub-succession of s_n
It is clear that every succession has at least one limit of oscillation
Proposition 2
- The upper limit of a succession
s_n is the maximum (in \mathbb{\overline{R}}) of its oscillation limits
- The lower bound of a succession
s_n is the minimum (in \mathbb{\overline{R}}) of its oscillation limits
Demonstration of Proposition 2 A)
Given the succession s_n suppose 3 cases:
Case 1: suppose \underset {n \to \infty} {\lim} \text{sup } s_n = s \in \mathbb{R}
Dice \forall n \in \mathbb{N}; x_n = \text{sup }\left \{ s_k|k\geqslant n \right \}; y_n = \text{inf }\left \{ s_k|k\geqslant n \right \}
By the definition of tiny, \exists N_1 \in \mathbb{N} such that if n > N_1 it is s-1 < x_n < s+1, and by the definition of supreme \exists \varphi(2)>\varphi(1) such that \frac{s-1}{2} < s_{\varphi(2)} < s + \frac{1}{2} and in general \exists \varphi(n) < \cdots < \varphi(1) such that \frac{s-1}{n} < s_{\varphi(n)} < s + \frac{1}{n}
When applying the Sandwich rule, the sub-succession s_{\varphi(n)} converges to s
Case 2: suppose \underset {n \to \infty} {\lim} \text{sup } s_n = +\infty and s_n is not growing
Dice \forall n \in \mathbb{N};x_n = +\infty
As a result, \forall n \in \mathbb{N}; \exists s_{\varphi(n)} such that \exists s_{\varphi(n)} > n
In addition it can be assumed that \exists \varphi(n) < \cdots < \varphi(1)
Therefore s_{\varphi(n)} \to +\infty
Case 3: suppose \underset {n \to \infty} {\lim} \text{sup } s_n = -\infty
When applying Proposition 1 C) it is evident
Demonstration of Proposition 2 B)
Given the succession s_n suppose 3 cases:
Case 1: suppose \underset {n \to \infty} {\lim} \text{inf } s_n = s \in \mathbb{R}
Dice \forall n \in \mathbb{N}; x_n = \text{sup }\left \{ s_k|k\geqslant n \right \}; y_n = \text{inf }\left \{ s_k|k\geqslant n \right \}
By the definition of tiny, \exists N_1 \in \mathbb{N} such that if n > N_1 it is s-1 < x_n < s+1, and by the definition of supreme \exists \varphi(2)>\varphi(1) such that \frac{s-1}{2} < s_{\varphi(2)} < s + \frac{1}{2} and in general \exists \varphi(n) < \cdots < \varphi(1) such that \frac{s-1}{n} < s_{\varphi(n)} < s + \frac{1}{n}
When applying the Sandwich rule, the sub-succession s_{\varphi(n)} converges to s
Case 2: suppose \underset {n \to \infty} {\lim} \text{inf } s_n = +\infty and s_n is not growing
Dice \forall n \in \mathbb{N};y_n = +\infty
As a result, \forall n \in \mathbb{N}; \exists s_{\varphi(n)} such that \exists s_{\varphi(n)} > n
In addition it can be assumed that \exists \varphi(n) < \cdots < \varphi(1)
Therefore s_{\varphi(n)} \to +\infty
Case 3: suppose \underset {n \to \infty} {\lim} \text{sup } s_n = -\infty
When applying Proposition 1 C) it is evident
Finally, it suffices to show that any other limit of oscillation of (s_n) is between the upper and lower bounds
It (s_{\varphi(n)}) convergent subsucession to a limit of oscillation
Using y_n \leq s_{\varphi(n)} \leq x_n and taking limits you get the result of proposition 2, which is what we wanted to prove
Proposition 3
Are (s_n) and (t_n) successions of real numbers and c\in \mathbb{R}
-
If \underset {n \to \infty} {\lim} \text{sup }s_n < c, \exists n_0|s_n < c, \forall n\geqslant n_0
That is, there is only a finite number of terms of the succession greater than or equal to c
-
If \underset {n \to \infty} {\lim} \text{sup }s_n > c, \exists n|s_n > c
That is, there is an infinite number of terms of the succession greater than c
-
If \underset {n \to \infty} {\lim} \text{inf }s_n > c, \exists n_0|s_n > c, \forall n\geqslant n_0
That is, there is only a finite number of terms in the succession less than or equal to c
-
If \underset {n \to \infty} {\lim} \text{inf }s_n < c, \exists n|s_n < c
That is, there are an infinite number of terms of the succession less than c
-
If \exists m|s_n < t_m, n > m \Rightarrow \left\{\begin{matrix}
\underset {n \to \infty} {\lim} \text{inf }s_n \leq \underset {n \to \infty} {\lim} \text{inf }t_n\\
\underset {n \to \infty} {\lim} \text{sup }s_n \leq \underset {n \to \infty} {\lim} \text{sup }t_n
\end{matrix}\right.
-
If \underset {n \to \infty} {\lim} \text{inf }s_n + \underset {n \to \infty} {\lim} \text{inf }t_n \leq \underset {n \to \infty} {\lim} \text{inf } (s_n + t_n) \leq \underset {n \to \infty} {\lim} \text{inf }s_n + \underset {n \to \infty} {\lim} \text{sup }t_n \leq \underset {n \to \infty} {\lim} \text{sup }s_n + \underset {n \to \infty} {\lim} \text{sup }t_n
-
If c\geq 0
\left\{\begin{matrix}
\underset {n \to \infty} {\lim} \text{inf } (c \cdot s_n) = c \cdot \underset {n \to \infty} {\lim} \text{inf } s_n\\
\underset {n \to \infty} {\lim} \text{sup } (c \cdot s_n) = c \cdot \underset {n \to \infty} {\lim} \text{sup } s_n
\end{matrix}\right.
-
If c\leq 0
\left\{\begin{matrix}
\underset {n \to \infty} {\lim} \text{inf } (c \cdot s_n) = c \cdot \underset {n \to \infty} {\lim} \text{sup } s_n\\
\underset {n \to \infty} {\lim} \text{sup } (c \cdot s_n) = c \cdot \underset {n \to \infty} {\lim} \text{inf } s_n
\end{matrix}\right.
-
If s_n \leq 0, t_n \leq 0
\underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{inf } t_n \leq \underset {n \to \infty} {\lim} \text{inf } (s_n \cdot t_n) \leq \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n \leq \underset {n \to \infty} {\lim} \text{sup } (s_n \cdot t_n) \leq \underset {n \to \infty} {\lim} \text{sup } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n
-
If s_n \leq 0, \forall n
\underset {n \to \infty} {\lim} \text{inf } \frac{s_{n+1}}{s_n} \leq \underset {n \to \infty} {\lim} \text{inf } \sqrt[n]{s_n} \leq \underset {n \to \infty} {\lim} \text{sup } \sqrt[n]{s_n} \leq \underset {n \to \infty} {\lim} \text{sup } \frac{s_{n+1}}{s_n}
Demonstration of A)
It \underset {n \to \infty} {\lim} \text{sup}_n s_n = l, then \forall \epsilon >0,\exists n_0\rightarrow \left \| \text{sup}_{k\geq n}s_k-l \right \|<\epsilon, for n\geq n_0
In particular, we need to \epsilon =c-l and therefore \text{sup}_{k\geq n}s_k>\epsilon +l=c, for n\geq n_0
Then s_n< c, for n\geq n_0 which is what we wanted to demonstrate
Demonstration of B)
Suppose that s_n> c only for a finite number of values of n and that n_1 is the maximum of such values
Then s_n< c, for n\geq n_1 and therefore, in that case \text{sup}_{k\geq n}s_k\leq c and \underset {n \to \infty}{\lim} \text{sup}_n sn \leq c
With what using the reduction to the absurd we arrive at what we wanted to demonstrate
Demonstration of C)
It \underset {n \to \infty} {\lim} \text{inf}_n s_n = l, then \forall \epsilon >0,\exists n_0\rightarrow \left \| \text{inf}_{k\geq n}s_k-l \right \|>\epsilon, for n\geq n_0
In particular, we need to \epsilon =c-l and therefore \text{inf}_{k\geq n}s_k<\epsilon +l=c, for n\geq n_0
Then s_n> c, for n\geq n_0 which is what we wanted to demonstrate
Demonstration of D)
Suppose that s_n< c only for a finite number of values of n and that n_1 is the minimum of such values
Then s_n> c, for n\geq n_1 and therefore, in that case \text{inf}_{k\geq n}s_k\geq c and \underset {n \to \infty}{\lim} \text{inf}_n sn \geq c
With what using the reduction to the absurd we arrive at what we wanted to demonstrate
Demonstration of E)
It \underset {n \to \infty} {\lim} \text{inf}_n s_n = l_s, then \forall \epsilon >0,\exists m\rightarrow \left \| \text{inf}_{k\geq n}s_k-l_s \right \|>\epsilon, for n>m
It \underset {n \to \infty} {\lim} \text{inf}_n t_n = l_t, then \forall \epsilon >0,\exists m\rightarrow \left \| \text{inf}_{k\geq n}t_k-l_t \right \|>\epsilon, for n>m
As s_n\leq t_n and if we apply C), then we have to \underset {n \to \infty} {\lim} \text{inf}_n s_n\leq \underset {n \to \infty} {\lim} \text{inf}_n t_n with what is proven that first part
It \underset {n \to \infty} {\lim} \text{sup}_n s_n = l_s, then \forall \epsilon >0,\exists m\rightarrow \left \| \text{sup}_{k\geq n}s_k-l_s \right \|<\epsilon, for n>m
It \underset {n \to \infty} {\lim} \text{sup}_n t_n = l_t, then \forall \epsilon >0,\exists m\rightarrow \left \| \text{sup}_{k\geq n}t_k-l_t \right \|<\epsilon, for n>m
As s_n\leq t_n and if we apply A), then we have to \underset {n \to \infty} {\lim} \text{sup}_n s_n\leq \underset {n \to \infty} {\lim} \text{sup}_n t_n with what is proven that second part
Demonstration of F)
Dice a, b \in \mathbb{\overline{R}} by definition, if \underset {n \to \infty} {\lim} s_n = a and \underset {n \to \infty} {\lim} t_n = b then (s_n + t_n) has limit a + b and therefore \underset {n \to \infty} {\lim} \left ( s_n + t_n \right ) = \left ( a + b \right )
As a \leq \left ( a + b \right ) and b \leq \left ( a + b \right ) then \left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_n\leq \underset {n \to \infty} {\lim} \left ( s_n + t_n \right ),\text{si } a\leq \left ( a + b \right )\\
\underset {n \to \infty} {\lim} t_n\leq \underset {n \to \infty} {\lim} \left ( s_n + t_n \right ),\text{si } b\leq \left ( a + b \right )
\end{matrix}\right.
From the definition of upper and lower limit we have to \underset {n \to \infty} {\lim} \text{inf } s_n \leq \underset {n \to \infty} {\lim} {\text{sup } } s_n and \underset {n \to \infty} {\lim} \text{inf } t_n \leq \underset {n \to \infty} {\lim} \text{sup } t_n
Then we can take \underset {n \to \infty} {\lim} s_n and \underset {n \to \infty} {\lim} t_n as lower limits of \underset {n \to \infty} {\lim} \left ( s_n + t_n \right )
And \underset {n \to \infty} {\lim} \left ( s_n + t_n \right ) as an upper limit of \underset {n \to \infty} {\lim} s_n and \underset {n \to \infty} {\lim} t_n
With what we have to \left\{\begin{matrix}
\underset {n \to \infty} {\lim} \text{inf } s_n \leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )\\
\underset {n \to \infty} {\lim} \text{inf } t_n \leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
\end{matrix}\right.
With what we have found the fourth inequality, let's continue with the demonstration
And if we take \underset {n \to \infty} {\lim} \text{inf } s_n and \underset {n \to \infty} {\lim} \text{inf } t_n as the sum of the lower limits of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
With what we have found the first and fourth inequalities, let's continue with the demonstration
And if we take \underset {n \to \infty} {\lim} \text{sup } s_n and \underset {n \to \infty} {\lim} \text{sup } t_n as the sum of the upper limits of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } s_n + \underset {n \to \infty} {\lim} \text{sup } t_n
With what we have found the first, the fourth and the fifth inequalities, let's continue with the demonstration
And if we take s_n + t_n as a lower limit of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } \left (s_n + t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
With what we have found the second inequality, let's continue with the demonstration
And if we take s_n + t_n as an upper limit of \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{inf } t_n
We have to \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{inf } \left ( s_n + t_n \right )
With what we have to \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{inf } \left ( s_n + t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } s_n + \underset {n \to \infty} {\lim} \text{sup } t_n
With what we have found the first, the second, the fourth and the fifth inequalities, let's continue with the demonstration
And if we take \underset {n \to \infty} {\lim} \text{sup } t_n as an upper limit of \underset {n \to \infty} {\lim} \text{inf } \left ( s_n + t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } \left ( s_n + t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } t_n
And if we take \underset {n \to \infty} {\lim} \text{inf } s_n as a lower limit of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } s_n\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
And if we take \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{sup } t_n as the sum of the upper limits of \underset {n \to \infty} {\lim} \text{inf } \left ( s_n + t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } \left ( s_n + t_n \right )\leq \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{sup } t_n
With what we have found the third inequality, let's continue with the demonstration
And if we take \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{sup } t_n as the sum of the lower limits of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{sup } t_n\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )
With what we have to \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{inf } \left ( s_n + t_n \right )\leq \underset {n \to \infty} {\lim} \text{inf } s_n + \underset {n \to \infty} {\lim} \text{sup } t_n\leq\underset {n \to \infty} {\lim} \text{sup } \left ( s_n + t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } s_n + \underset {n \to \infty} {\lim} \text{sup } t_n
With what we have found all the inequalities, with which we end the demonstration
Demonstration of G)
Dice a, b, c \in \mathbb{\overline{R}} by definition, if \underset {n \to \infty} {\lim} s_n = a and \underset {n \to \infty} {\lim} t_n = b then if (c \cdot s_n) is convergent, has limit c \cdot a
Therefore \underset {n \to \infty} {\lim} \left ( c \cdot s_n \right ) = \left ( c \cdot a \right )
As (c \cdot s_n) it must be convergent and c\geq 0, we have to \underset {n \to \infty} {\lim} \text{inf } \left ( c \cdot s_n \right ) = \underset {n \to \infty} {\lim} \text{sup } \left ( c \cdot s_n \right ) = \left ( c \cdot a \right )
Replace \underset {n \to \infty} {\lim} \text{inf } \left ( c \cdot s_n \right ) = \left ( c \cdot a \right ) in the equality that we want to prove and we have to \left ( c \cdot a \right ) = c\cdot \underset {n \to \infty} {\lim} \text{inf } s_n
As \underset {n \to \infty} {\lim} s_n it is bounded we have to \underset {n \to \infty} {\lim} \text{inf } s_n = \underset {n \to \infty} {\lim} \text{sup } s_n = a
Replace \underset {n \to \infty} {\lim} \text{inf } s_n in the equality that we want to prove and we have to \left ( c \cdot a \right ) = \left ( c \cdot a \right ) which demonstrates the first equality
Replace \underset {n \to \infty} {\lim} \text{sup } \left ( c \cdot s_n \right ) = \left ( c \cdot a \right ) in the equality that we want to prove and we have to \left ( c \cdot a \right ) = c\cdot \underset {n \to \infty} {\lim} \text{sup } s_n
Replace \underset {n \to \infty} {\lim} \text{sup } s_n in the equality that we want to prove and we have to \left ( c \cdot a \right ) = \left ( c \cdot a \right ) with which the second equality is demonstrated, ending the demonstration
Demonstration of H)
Dice a, b, c \in \mathbb{\overline{R}} by definition, if \underset {n \to \infty} {\lim} s_n = a and \underset {n \to \infty} {\lim} t_n = b then if (c \cdot s_n) is convergent, has limit c \cdot a
Therefore \underset {n \to \infty} {\lim
} \left ( c \cdot s_n \right ) = \left ( c \cdot a \right )
As (c \cdot s_n) it must be convergent and c < 0, we have to \underset {n \to \infty} {\lim} \text{inf } \left ( c \cdot s_n \right ) = \underset {n \to \infty} {\lim} \text{sup } \left ( c \cdot s_n \right ) = \left ( c \cdot a \right )
Replace \underset {n \to \infty} {\lim} \text{inf } \left ( c \cdot s_n \right ) = \left ( c \cdot a \right ) in the equality that we want to prove and we have to \left ( c \cdot a \right ) = c\cdot \underset {n \to \infty} {\lim} \text{sup } s_n
As \underset {n \to \infty} {\lim} s_n it is bounded we have to \underset {n \to \infty} {\lim} \text{inf } s_n = \underset {n \to \infty} {\lim} \text{sup } s_n = a
Replace \underset {n \to \infty} {\lim} \text{sup } s_n in the equality that we want to prove and we have to \left ( c \cdot a \right ) = \left ( c \cdot a \right ) which demonstrates the first equality
Replace \underset {n \to \infty} {\lim} \text{sup } \left ( c \cdot s_n \right ) = \left ( c \cdot a \right ) in the equality that we want to prove and we have to \left ( c \cdot a \right ) = c\cdot \underset {n \to \infty} {\lim} \text{inf } s_n
Replace \underset {n \to \infty} {\lim} \text{inf } s_n in the equality that we want to prove and we have to \left ( c \cdot a \right ) = \left ( c \cdot a \right ) with which the second equality is demonstrated, ending the demonstration
Demonstration of I)
Dice a, b \in \mathbb{\overline{R}} by definition, if \underset {n \to \infty} {\lim} s_n = a and \underset {n \to \infty} {\lim} t_n = b then (s_n \cdot t_n) has limit a \cdot b and therefore \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right ) = \left ( a \cdot b \right )
In addition s_n\geq 0 and t_n\geq 0
As a \leq \left ( a \cdot b \right ) and b \leq \left ( a \cdot b \right ) then \left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_n\leq \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right ),\text{si } a\leq \left ( a \cdot b \right )\\
\underset {n \to \infty} {\lim} t_n\leq \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right ),\text{si } b\leq \left ( a \cdot b \right )
\end{matrix}\right.
From the definition of upper and lower limit we have to \underset {n \to \infty} {\lim} \text{inf } s_n \leq \underset {n \to \infty} {\lim} {\text{sup } } s_n and \underset {n \to \infty} {\lim} \text{inf } t_n \leq \underset {n \to \infty} {\lim} \text{sup } t_n
Then we can take \underset {n \to \infty} {\lim} s_n and \underset {n \to \infty} {\lim} t_n as lower limits of \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right )
And \underset {n \to \infty} {\lim} \left ( s_n \cdot t_n \right ) as an upper limit of \underset {n \to \infty} {\lim} s_n and \underset {n \to \infty} {\lim} t_n
With what we have to \left\{\begin{matrix}
\underset {n \to \infty} {\lim} \text{inf } s_n \leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )\\
\underset {n \to \infty} {\lim} \text{inf } t_n \leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
\end{matrix}\right.
With what we have found the fourth inequality, let's continue with the demonstration
And if we take \underset {n \to \infty} {\lim} \text{inf } s_n and \underset {n \to \infty} {\lim} \text{inf } t_n as the product of the lower limits of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
With what we have found the first and fourth inequalities, let's continue with the demonstration
And if we take \underset {n \to \infty} {\lim} \text{sup } s_n and \underset {n \to \infty} {\lim} \text{sup } t_n as the product of the upper limits of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n
With what we have found the first, the fourth and the fifth inequalities, let's continue with the demonstration
And if we take s_n \cdot t_n as a lower limit of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } \left (s_n \cdot t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
With what we have found the second inequality, let's continue with the demonstration
And if we take s_n \cdot t_n as an upper limit of \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{inf } t_n
We have to \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{inf } \left ( s_n \cdot t_n \right )
With what we have to \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{inf } \left ( s_n \cdot t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n
With what we have found the first, the second, the fourth and the fifth inequalities, let's continue with the demonstration
And if we take \underset {n \to \infty} {\lim} \text{sup } t_n as an upper limit of \underset {n \to \infty} {\lim} \text{inf } \left ( s_n \cdot t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } \left ( s_n \cdot t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } t_n
And if we take \underset {n \to \infty} {\lim} \text{inf } s_n as a lower limit of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } s_n\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
And if we take \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n as a product of the upper limits of \underset {n \to \infty} {\lim} \text{inf } \left ( s_n \cdot t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } \left ( s_n \cdot t_n \right )\leq \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n
With what we have found the third inequality, let's continue with the demonstration
And if we take \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n as the product of the lower limits of \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
We have to \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n\leq \underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )
With what we have to \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{inf } t_n\leq \underset {n \to \infty} {\lim} \text{inf } \left ( s_n \cdot t_n \right )\leq \underset {n \to \infty} {\lim} \text{inf } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n\leq\underset {n \to \infty} {\lim} \text{sup } \left ( s_n \cdot t_n \right )\leq \underset {n \to \infty} {\lim} \text{sup } s_n \cdot \underset {n \to \infty} {\lim} \text{sup } t_n
With what we have found all the inequalities, with which we end the demonstration
Demonstration of J)
Dice a, b \in \mathbb{\overline{R}} and b\neq 0by definition, if \underset {n \to \infty} {\lim} s_{n+1} = a and \underset {n \to \infty} {\lim} s_n = b then \left ( \frac{s_{n+1}}{s_n} \right ) has limit \frac{a}{b} and therefore \underset {n \to \infty} {\lim} \left ( \frac{s_{n+1}}{s_n} \right ) = \left ( \frac{a}{b} \right )
In addition s_n > 0
As a \leq \left ( \frac{a}{b} \right ) and b \leq \left ( \frac{a}{b} \right ) then \left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_{n+1}\leq \underset {n \to \infty} {\lim} \left ( \frac{s_{n+1}}{s_n} \right ),\text{si } a\leq \left ( \frac{a}{b} \right )\\
\underset {n \to \infty} {\lim} s_n\leq \underset {n \to \infty} {\lim} \left ( \frac{s_{n+1}}{s_n} \right ),\text{si } b\leq \left ( \frac{a}{b} \right )
\end{matrix}\right.
From the definition of upper and lower limit we have to \underset {n \to \infty} {\lim} \text{inf } s_{n+1} \leq \underset {n \to \infty} {\lim} {\text{sup } } s_{n+1} and \underset {n \to \infty} {\lim} \text{inf } s_n \leq \underset {n \to \infty} {\lim} \text{sup } s_n
Then we can take \underset {n \to \infty} {\lim} s_{n+1} and \underset {n \to \infty} {\lim} s_n as lower limits of \underset {n \to \infty} {\lim} \left ( \frac{s_{n+1}}{s_n} \right )
And \underset {n \to \infty} {\lim} \left ( \frac{s_{n+1}}{s_n} \right ) as an upper limit of \underset {n \to \infty} {\lim} s_{n+1} and \underset {n \to \infty} {\lim} s_n
With what we have to \left\{\begin{matrix}
\underset {n \to \infty} {\lim} \text{inf } s_{n+1} \leq \underset {n \to \infty} {\lim} \text{sup } \left ( \frac{s_{n+1}}{s_n} \right )\\
\underset {n \to \infty} {\lim} \text{inf } s_n \leq \underset {n \to \infty} {\lim} \text{sup } \left ( \frac{s_{n+1}}{s_n} \right )
\end{matrix}\right.
With what we have found the fourth inequality, let's continue with the demonstration
And if we take \frac{s_{n+1}}{s_n} as a lower limit of \underset {n \to \infty} {\lim} \text{sup } \left ( \frac{s_{n+1}}{s_n} \right )
We have to \underset {n \to \infty} {\lim} \text{inf } \left (\frac{s_{n+1}}{s_n} \right )\leq \underset {n \to \infty} {\lim} \text{sup } \left ( \frac{s_{n+1}}{s_n} \right )
With what we have found the first and fourth inequalities, let's continue with the demonstration
Suppose that b > 0
We take \alpha such that 0 < \alpha < s, by C) we have to \exists n_0 | \frac{s_{k+1}}{s_k} > \alpha, k\geq n_0
We multiply these inequalities for each n > n_0 with the values k = n_0, n_0 + 1, \cdots n - 2, n - 1
Getting \frac{s_n}{s_{n_0}} > \alpha^{n-n_0}, n\geq n_0
Which is equivalent to \sqrt[n]{s_n} > \alpha \cdot \left ( s_{n_0}\cdot \alpha^{-N} \right )^{\frac{1}{n}}, n\geq n_0
And that implies that \underset {n \to \infty} {\lim} \text{inf } \sqrt[n]{s_n} > \alpha \cdot \underset {n \to \infty} {\lim} \text{inf } \left ( s_{n_0}\cdot \alpha^{-N} \right )^{\frac{1}{n}} = \alpha \cdot \underset {n \to \infty} {\lim} \left ( s_{n_0}\cdot \alpha^{-N} \right )^{\frac{1}{n}} = \alpha
Since \underset {n \to \infty} {\lim} \left ( s_{n_0}\cdot \alpha^{-N} \right )^{\frac{1}{n}} = 1
As \alpha \leq b and b = \underset {n \to \infty} {\lim} \text{inf } \frac{s_{n+1}}{s_n} then \underset {n \to \infty} {\lim} \text{inf } \frac{s_{n+1}}{s_n} \leq \underset {n \to \infty} {\lim} \text{inf } \sqrt[n]{s_n}
With what we have found the second inequality, let's continue with the demonstration
We have to \underset {n \to \infty} {\lim} \text{inf } \frac{s_{n+1}}{s_n} \leq \underset {n \to \infty} {\lim} \text{inf } \sqrt[n]{s_n} \leq \underset {n \to \infty} {\lim} \text{sup } \frac{s_{n+1}}{s_n}
Then we can take \underset {n \to \infty} {\lim} \sqrt[n]{s_n} as an upper limit of \underset {n \to \infty} {\lim} \text{inf } \frac{s_{n+1}}{s_n} and \underset {n \to \infty} {\lim} \text{inf } \sqrt[n]{s_n}
With what we have to \underset {n \to \infty} {\lim} \text{inf } \frac{s_{n+1}}{s_n} \leq \underset {n \to \infty} {\lim} \text{inf } \sqrt[n]{s_n} \leq \underset {n \to \infty} {\lim} \text{sup } \sqrt[n]{s_n} \leq \underset {n \to \infty} {\lim} \text{sup } \frac{s_{n+1}}{s_n}
With what we have found all the inequalities, with which we end the demonstration
Limit of succession and elementary functions
If f(x) represents an elementary function (e^x, \log x, \cos x, \tan x, \arcsin x, \arccos x, \arctan x, x^{y}), then \underset {n \to \infty} {\lim} s_n = a \Rightarrow \underset {n \to \infty} {\lim} (s_n) = f(a) for any point a of the domain of the function and any succession s_n contained in the domain of the function
Other elementary limits that we must know are:
\underset {n \to \infty} {\lim} s_n = -\infty \Rightarrow \underset {n \to \infty} {\lim} e^{s_n} = 0
\underset {n \to \infty} {\lim} s_n = +\infty \Rightarrow \underset {n \to \infty} {\lim} e^{s_n} = +\infty
\underset {n \to \infty} {\lim} s_n = 0, s_n > 0, \forall n \Rightarrow \underset {n \to \infty} {\lim} \log s_n = -\infty
\underset {n \to \infty} {\lim} s_n = +\infty \Rightarrow \underset {n \to \infty} {\lim} \log s_n = +\infty
\underset {n \to \infty} {\lim} s_n = -\infty \Rightarrow \underset {n \to \infty} {\lim} \arctan s_n = -\frac{\pi}{2}
\underset {n \to \infty} {\lim} s_n = +\infty \Rightarrow \underset {n \to \infty} {\lim} \arctan s_n = \frac{\pi}{2}
\underset {n \to \infty} {\lim} s_n = 0, s_n > 0, \forall n \Rightarrow \underset {n \to \infty} {\lim} s_{n}^{r} = \left\{\begin{matrix}
0, \text{si } r > 0\\
+\infty, \text{si } r < 0
\end{matrix}\right.
\underset {n \to \infty} {\lim} s_n = +\infty\Rightarrow \underset {n \to \infty} {\lim} s_{n}^{r} = \left\{\begin{matrix}
+\infty, \text{si } r > 0\\
0, \text{si } r < 0
\end{matrix}\right.
- If
f(x) = a_r \cdot x^r + a_{r - 1} \cdot x^{r - 1} + \cdots + a_0 is a polynomial with r \in \mathbb{N}, a_r \neq 0 then \left\{\begin{matrix}
\underset {n \to \infty} {\lim} s_n = +\infty \Rightarrow \underset {n \to \infty} {\lim} f(s_n) = +\infty, \text{si } a_r > 0\\
\underset {n \to \infty} {\lim} s_n = +\infty \Rightarrow \underset {n \to \infty} {\lim} f(s_n) = -\infty, \text{si } r < 0
\end{matrix}\right.
Equivalences
Are (s_n) and (t_n) successions we will say that they are equivalents if it is true that:
\underset {n \to \infty} {\lim} \frac{s_n}{t_n} = 1
And we'll denote it as s_n \sim t_n
Main equivalences
- If
s_n \to 0 then \left\{\begin{matrix}
e^{s_n} - 1 \sim s_n \\
\log {(1 + s_n)} \sim s_n \\
\sin s_n \sim s_n \\
1 - \cos s_n\sim \frac{1}{2} \cdot s_{n}^{2}
\end{matrix}\right.
- If
f(x) = a_r \cdot x^r + a_{r - 1} \cdot x^{r - 1} + \cdots + a_0 is a polynomial with r \in \mathbb{N}, a_r \neq 0 and s_n\to +\infty then \left\{\begin{matrix}
f(x) \sim a_r \cdot s_{n}^{r}\\
\log (f(s_n)) \sim r \cdot \log (s_n), \text{si } a_r > 0
\end{matrix}\right.
- Stirling's formula
n! \sim n^{n} \cdot e^{-n} \cdot \sqrt{2 \cdot \pi \cdot n}
Wallis Product
The Wallis product has the following form:
\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots=\frac{\pi}{2}
Demonstration of Wallis Product
We can express it as the following succession with s_1, s_n = \frac{3}{2}\cdot\frac{5}{4}\cdots\frac{2\cdot n - 1}{2\cdot n - 1}, n\geq 2
If we take the partial products of the Wallis product with an odd number of factors o_n=\frac{2^2\cdot4^2\cdots\left ( 2\cdot n - 2 \right )^2\cdot\left ( 2\cdot n\right )}{1\cdot3^2\cdots\left ( 2\cdot n - 2 \right )^2} and with even number of factors e_n=\frac{2^2\cdot4^2\cdots\left ( 2\cdot n - 2 \right )^2}{1\cdot3^2\cdots\left ( 2\cdot n - 3 \right )^2\cdots\left ( 2\cdot n - 1 \right )} with e_1 = 1 we interpret it as an empty product
We need to:
o_n = \frac{2\cdot n}{s_{n}^{2}} and e_n = \frac{2\cdot n - 1}{s_{n}^{2}}
It is evident that e_n < e_{n+1}, o_n <o_{n+1} and comparing it to o_n = \frac{2\cdot n}{s_{n}^{2}} and e_n = \frac{2\cdot n - 1}{s_{n}^{2}} we have to:
e_1 < e_2 < e_3 < \cdots < o_3 <o_2 < o_1
If 1\leq i\leq n, o_n\leq o_i=\frac{2\cdot i}{s_{i}^{2}} and e_n\geq e_i = \frac{2\cdot i - 1}{s_{i}^{2}} we deduce that \frac{2\cdot i - 1}{e_n}\leq s_{i}^{2}\leq \frac{2\cdot i}{o_n}
If we define the succession a_0 = 1 and a_n = s_{n+1}-s_n with n\geq 1
It is true that a_n = s_{n+1}-s_n=s_n\cdot \left ( \frac{2\cdot n + 1}{2\cdot n} \right )=\frac{s_n}{2\cdot n}=\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2\cdot n-1}{2\cdot n}\cdots
As a_i\cdot a_j=\frac{j+1}{i+j+1} \cdot a_i\cdot a_{j+1}+\frac{i+1}{i+j+1} \cdot a_{i+1}\cdot a_j, we have to a_{k+1}=\frac{2\cdot k+1}{2\cdot(k+1)} \cdot a_k and \frac{j+1}{i+j+1}\cdot a_i \cdot a_{j+1}+\frac{i+1}{i+j+1}\cdot a_{i+1} \cdot a_j=a_i \cdot a_j\cdot \left ( \frac{j+1}{i+j+1}\cdot\frac{2\cdot j+1}{2\cdot \left ( j+1 \right )} + \frac{i+1}{i+j+1}\cdot\frac{2\cdot i+1}{2\cdot \left ( i+1 \right )} \right )=a_i \cdot a_j
So a_i\cdot a_j=\frac{j+1}{i+j+1} \cdot a_i\cdot a_{j+1}+\frac{i+1}{i+j+1} \cdot a_{i+1}\cdot a_j is demonstrated
The fundamental property of succession a_n sets that a_0\cdot a_n+a_1\cdot a_{n-1}+\cdots+a_n\cdot a_0=1
Which we're going to test by induction
It is fulfilled for a_{0}^{2}
Let's test if it's true n+1 if met for n
Apply a_i\cdot a_j=\frac{j+1}{i+j+1} \cdot a_i\cdot a_{j+1}+\frac{i+1}{i+j+1} \cdot a_{i+1}\cdot a_j to each summand to obtain:
1=a_0\cdot a_n+a_1\cdot a_{n-1}+\cdots+a_n\cdot a_0=\left ( a_0\cdot a_{n+1}+\frac{1}{n+1}\cdot a_1\cdot a_n \right )+\left ( \frac{n}{n+1} cdot a_1\cdot a_n +\frac{2}{n}\cdot a_2\cdot a_{n-1}\right )+\cdots+\left ( \frac{1}{n+1}\cdot a_n\cdot a_1+a_{n+1}\cdot a_0 \right )=a_0\cdot a_{n+1}+a_1\cdot a_n+\cdots+a_{n+1}\cdot a_0 with what is demonstrated
Now suppose we have the first quadrant of a coordinate system divided into rectangles by the lines x=s_n and y=s_n
It R_{i,j} the rectangle whose lower-left and upper-right corners (s_i,s_j) and (s_i+1,s_j+1)
The area of each rectangle a_i\cdot a_j and therefore identity a_0\cdot a_n+a_1\cdot a_{n-1}+\cdots+a_n\cdot a_0=1 tells us that the sum of the areas of the rectangles R_{i,j} such that i+j=0 is 1
Let us denote P_n as the polygonal region formed by the rectangles R_{i,j} such that i+j<n
The outer corners of P_n are the points (s_i,s_j) for which i+j=n+1, with 1\leq i,j\leq n, and the distance from the coordinate origin to each of them is \sqrt{s_{i}^{2}+s_{j}^{2}}
If we apply \frac{2\cdot i - 1}{e_n}\leq s_{i}^{2}\leq \frac{2\cdot i}{o_n}, each of these distances shall be bounded above by \sqrt{\frac{2\cdot\left ( i+j \right )}{o_n}}=\sqrt{\frac{2\cdot\left ( n+1 \right )}{o_n}}
Similarly, the inner corners of P_n are the points (s_i,s_j) for which i+j=n, with 0\leq i,j\leq n, and the distance from the coordinate origin to each of them is bounded below, using again \frac{2\cdot i - 1}{e_n}\leq s_{i}^{2}\leq \frac{2\cdot i}{o_n} we have to \sqrt{\frac{2\cdot\left ( i+j-1 \right )}{e_n}}=\sqrt{\frac{2\cdot\left ( n-1 \right )}{e_n}}
Therefore, each polygon P_n is contained in a quarter radius circumference \sqrt{\frac{2\cdot\left ( n+1 \right )}{o_n}} and contains a quarter radius circumference \sqrt{\frac{2\cdot\left ( n-1 \right )}{e_n}}
Then, using that area of P_n is n, we have to \frac{n-1}{e_n}\cdot \frac{\pi}{2}<e_n<o_n<\frac{n+1}{n}\cdot \frac{\pi}{2}
And how \underset {n \to \infty} {\lim}e_n=\underset {n \to \infty} {\lim}o_n=\frac{\pi}{2} we have to \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots=\frac{\pi}{2} thus ending the demonstration
Stirling's formula
Stirling's formula has the following form:
n! \sim n^{n} \cdot e^{-n} \cdot \sqrt{2 \cdot \pi \cdot n}
Demonstration of Stirling's Formula
Let's try to prove Stirling's formula using the Wallis product we just demonstrated
Specifically that \underset {n \to \infty} {\lim}\frac{n!}{n^{n} \cdot e^{-n} \cdot \sqrt{2 \cdot \pi \cdot n}}=1
For this we will use the expression x_n=\frac{n!}{n^{n} \cdot e^{-n} \cdot \sqrt{n}}
We have chosen this succession because a_n=\left ( 1+\frac{1}{n} \right )^n it is growing and bounded; therefore it has limit and is the number e
If we consider the succession b_n=\left ( 1+\frac{1}{n} \right )^{n+1} we get that the closed intervals I_n=\left \{ a_n,b_n \right \} verify that I_{n+1}\subset I_n and \underset {n \to \infty} {\lim}b_n-a_n=0
By the principle of embedded intervals, we know that both successions have the same limit, the number e and in addition a_i<e<b_i with i\geq 1
If we replace the previous inequalities for i=1,\cdots,n-1 we have to e\cdot\left ( \frac{n}{e} \right )^n<n!<e\cdot n\cdot\left ( \frac{n}{e} \right )^n
The choice of n^{n} \cdot e^{-n} \cdot \sqrt{n} was correct as it is an intermediate value between e\cdot\left ( \frac{n}{e} \right )^n and e\cdot n\cdot\left ( \frac{n}{e} \right )^n
Now let's see that x_n it is convergent, for this we take x_{n+1}<x_n and that x_n it is positive, we will have to \underset {n \to \infty} {\lim}x_n=\sigma with \exists \sigma\in\mathbb{R}
Then x_{n+1}<x_n\Leftrightarrow \frac{x_n}{x_{n+1}}>1\Leftrightarrow \frac{1}{e}\cdot\left ( 1+\frac{1}{n} \right )^{n+\frac{1}{2}}>1\Leftrightarrow \left ( n+\frac{1}{2} \right )\cdot \log{\left ( 1+\frac{1}{n} \right )}-1<0\Leftrightarrow \log{\left ( 1+\frac{1}{n} \right )}>\frac{1}{n+\frac{1}{2}}
The area bounded by the curve y=\frac{1}{x} and the OX axis between y=n and x=n+1 it is precisely \log{\left ( 1+\frac{1}{n} \right )}, this area is greater than the area of region A, bounded by the OX axis and the line tangent to the function y=\frac{1}{x} at the pointy=n++\frac{1}{2} between x=n and x=n+1
That is to say \log {\left ( 1+\frac{1}{n} \right )}>\acute{A}\text{rea}(A)=\frac{1}{2}\cdot\left ( \frac{n}{\left ( n+\frac{1}{2} \right )^2}+\frac{n+1}{\left ( n+\frac{1}{2} \right )^2} \right )=\frac{1}{n+\frac{1}{2}} so it is convergent
Since x_n is convergent, each sub-succession of x_nwill also be convergent and will have the same limit
Therefore \underset {n \to \infty} {\lim}x_n=\underset {n \to \infty} {\lim}\frac{x_n^2}{x_{2\cdot n}}=\sigma
With what we have to \frac{x_n^2}{x_{2\cdot n}}=\frac{\left ( 2\cdot n \right )^{2\cdot n}\cdot e^{-2\cdot n}\cdot\sqrt{2\cdot n}}{\left ( 2\cdot n \right )!}\cdot \frac{\left ( n! \right )^2}{n^{2\cdot n}\cdot e^{-2\cdot n}\cdot n}=\frac{2\cdot4\cdots (2\cdot n)}{1\cdot 3\cdots\left ( 2\cdot n-1 \right )}\cdot\frac{\sqrt{2}}{\sqrt{n}} \to \sqrt{2\cdot \pi} =\sigma with what we have arrived at the Wallis product that we have already demonstrated before and with which we have finished the demonstration
