Successions

Think that are the inheritance

Informally, a succession of real numbers is a list limited by numbers $s_1, s_2, s_3, s_4, \cdots, s_n, \cdots$ (where n indicates the place in the number $s_n$ list); it's obvious that this is a real function with domain $\mathbb{N}$

Succession

A succession of elements of a set is an application domain $\mathbb{N}$ and codomain said set. In particular, a sequence of real numbers is a real function with domain $\mathbb{R}$, that is, an application $s|\mathbb{N}\rightarrow\mathbb{R}$

Traditionally, the value that a succession takes in each $n\in\mathbb{N}$ it is denoted by $s_n$, instead of $s(n)$ like any other function. We will usually refer to $s_n$ with the name of the nth term of a succession, but it should not be overlooked that each term carries a double information: its value and the n place it occupies

As the domain $\mathbb{N}$ is common to all successions, rather than using notation $s|\mathbb{N}\rightarrow\mathbb{R}$, is more common to find notations of the type $(s_n)_{n\in\mathbb{N}}$ or $(s_n)_{n=1}^{\infty}\text{, }\left\{s_n\right\}_{n=1}^{\infty}$ or $(s_n)$, if it does not lead to confusion, or similar, with greater emphasis on the terms

Although notation may lead to confusion, it should not be necessary to insist on the difference between succession and the set of values that the succession takes, which is the same as between any function and its set of values (image set or range); note, for example, that a succession always has infinite terms even if it takes a single value, as is the case with constant successions

Successions are indicated by giving a formula that defines the term nth, the most common being:

• Constant succession:
  $s_n=a$, where a is a prefixed real number and consists of the terms $a, a, a, a, \cdots, a, \cdots$
• Succession of natural numbers:
  $s_n=n$, consists of the terms $1, 2, 3, 4, \cdots, n, \cdots$
• Succession of fractional numbers:
  $s_n=\frac{1}{n}$, consists of the terms $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \cdots, \frac{1}{n}, \cdots$
• Succession of 1, -1:
  $s_n=(-1)^n$, consists of the terms $-1, 1, -1, 1, \cdots, (-1)^n, \cdots$
• Complex algebraic successions:
  Formulas do not have to refer only to simple algebraic operations. For example, the succession of decimal approximations of $\pi$, consists of the terms $3.1, 3.14, 3.141, 3.1415, \cdots, 3.14159265, \cdots$

  We can give an explicit formula for the nth term with the help of the entire part function, even if we did not know how to write all the figures of the term, for example, a million

  Specifically, for each $n\in\mathbb{N}$, $s_n= 3 + \frac{a_1}{10} + \frac{a_2}{10^2} + \cdots + \frac{a_k}{10^k} + \cdots + \frac{a_n}{10^n}$ where $a_k= [10^k \cdot \pi] - 10 \cdot [10^{k-1} \cdot \pi] (1 \leq k\leq n)$

  The fact that this formula does not provide a computational algorithm for the $a_k$ it does not prevent that these are defined without ambiguity and without any exception

• Recurring successions:
  This name is given successions whose terms are defined according to the above (by means of an inductive or recursive definition). An example of this type is the succession of Fibonacci:

  $s_1=s_2=1, s_{n+2}=s_{n+1}+s_n, n\in\mathbb{N}$, consists of the terms $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, \cdots$

• Succession of non-mathematical
  The rule that defines a succession does not have to be of a strictly mathematical character. For example, we can define a succession as follows:

  $\tiny s_n=\left\{\begin{matrix}\frac{10^7}{3} & \text{if the name of the number n contains the letter d} \\ \sqrt{n} & \text{otherwise}\end{matrix}\right.$
  
  Or any other condition allowing to ensure that for each $n\in\mathbb{N}$ is associated with it without exception, unequivocally a single perfectly defined real number

• Succession, which is exactly a set numerical
  There are successions whose rank is exactly $\mathbb{Z}$ or $\mathbb{Q}$; the usual construction is done through The Diagonal Process of Cantor

  $s_1=0, s_2=1, s_3=\frac{1}{2}, s_4=\frac{-1}{2}, \cdots$

  This construction, which supports repeated, allows to prove that the whole of the rationals is numerable; that is, its cardinal matches that of the natural

Limit of the succession

A succession $(s_n)$ it is said convergent if there is a real number such that for each $\epsilon > 0$ you can find a natural number $N=N(\epsilon)$ so as long as $n>N$ check $\left \|s_n-a\right \|<\epsilon$

It is then said that the number is limit of the succession $(s_n)$ and writes $a=\underset {n \to \infty} {\lim} s_n$. We'll also say that $s_n$ converges the number to

The expression $s_n \to a$ is used to indicate that the sequence of nth term $s_n$ is convergent and has the limit

Remember that the inequality $\left \|s_n-a\right \|<\epsilon$ it is equivalent to the two inequalities $-\epsilon < s_n - a < \epsilon$ that are equivalent to the inequalities $a-\epsilon < s_n < a+\epsilon$

• The constant succession $s_n=c (c\in\mathbb{R})$ converges to the number $\epsilon$
• The succession $s_n=\frac{1}{n}$ converges to 0 (as a result of The Archemedian property)
• The succession $s_n=(-1)^n$ is not convergent if I had limit 1, taking $\epsilon=2$ in defining limit, it would have to be $\left \|s_n-1\right \|<2$ for all n big enough; However $\left \|s_n-1\right \|=2$ for all odd n'n. And if I had a limit $a\not=1$, taking $\epsilon=\left \|1-a\right \|$, it would have to be $\left \|s_n-a \right \| < \left \|1-a \right \|$ for all n big enough; However $\left \|s_n-a \right \| = \left \|1-a \right \|$ for all n pairs

  Conclusion: the succession has no limit

• The succession $s_n=n$ can't be convergent, because if I had a limit to, taking $\epsilon=1$ in the definition of convergence, for some N would have to be $n<a+1$ as long as n was greater than N, which is impossible (as a result of Archimadian property)

Proposition: uniqueness of the boundary of a succession

It $(s_n)$ a succession convergent and $a, b \in \mathbb{R}$ such that $a=\underset {n \to \infty} {\lim} s_n, b=\underset {n \to \infty} {\lim} s_n$ then $a=b$

Demonstration

Since a and b are the limits of the succession $s_n$Dice $\epsilon > 0$ there will be N and N' such that

$\left\{\begin{matrix}\left \| s_n-a \right \|\leq \frac{\epsilon}{2} & \text{if }n > N \\\left \| s_n-b \right \|\leq \frac{\epsilon}{2} & \text{if }n > N'\end{matrix}\right.$


Then $\left \|a-b\right\|=\left\|a-s_n+s_n-b\right\|\leq \left\|s_n-a\right\|+\left\|s_n-b\right\|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$, therefore $\left\|a-b\right\|=0\rightarrow a=b$

Conclusion: the limit of a sequence converging is the unique real number to which the succession converges

Succession bounded

A succession $(s_n)_{n=1}^{\infty}$ it is said that is bounded superiorly if there is a number $C\in\mathbb{R}$ such that for all $n\in\mathbb{N}, s_n \leq C$

It is said that is bounded inferiorly if there is a number $K\in\mathbb{R}$ such that for all $n\in\mathbb{N}, K\leq s_n$

It is said that is limited if it's top and bottom. It is equivalent to saying that there is a number $M\geq 0$ such that for all $n\in\mathbb{N}, \left\|s_n\right\|\leq M$

Proposition: Convergent and bounded

The whole succession is convergent it is bounded

Demonstration

It $(s_n)$ a sequence converging to a number $a\in\mathbb{R}$

We take $\epsilon=1$ in the definition of limit, and there will be some $N\in\mathbb{N}$ such that $\left\|s_n-a\right\| < 1$ for all $n >N$

If $B=\max\lbrace 1,\left\|s_1-a\right\|, \left\|s_2-a\right\|, \cdots, \left\|s_N-a\right\|\rbrace$ you have to $\left\|s_n-a\right\|\leq B$, that is to say, $a-B\leq s_n \leq a+B$ for all $n\in \mathbb{N}$

Conclusion: the succession is bounded

Succession monotonous

A succession $(s_n)$ it is monotonically non decreasing if $\forall n \in \mathbb{N}$ verified $s_n\leq s_{n+1}$

A succession $(s_n)$ it is monotonically non-increasing if $\forall n \in \mathbb{N}$ verified $s_n \geq s_{n+1}$

A succession $(s_n)$ it is strictly increasing if $\forall n \in \mathbb{N}$ verified $s_n < s_{n+1}$

A succession $(s_n)$ it is strictly decreasing if $\forall n \in \mathbb{N}$ verified $s_n > s_{n+1}$

A succession is monotonous if you meet any of the above cases

Proposition: equivalence with convergence at 0

If $(s_n)$ it is a succession bounded and $(t_n)$ is a convergent succession to 0, succession $(s_n \cdot t_n)$ converges to 0

Demonstration

It $\epsilon > 0, K > 0$ such that $\left\|s_n\right\| \leq K, \forall n \in\mathbb{N}$

Using the definition of convergence of $t_n$ for $\frac{\epsilon}{K}$ you have to $\left\|s_n \cdot t_n\right\| \leq K \cdot \left\|t_n\right\|\leq K\cdot \frac{\epsilon}{K}=\epsilon$

Conclusion: $(s_n \cdot t_n)$ converges to 0

Proposition: convergent if lying above or below

• It $(s_n)$ a non-decreasing monotonous succession. Then $(s_n)$ is convergent if and only if it is lying down higher, in which case $\underset {n \to \infty} {\lim} s_n=\sup\lbrace s_n|n\in\mathbb{N}\rbrace$
• It $(s_n)$ a monotonous non-increasing succession. So $(s_n)$ is convergent if and only if it is bounded below, in which case $\underset {n \to \infty} {\lim} s_n=\inf\lbrace s_n|n\in\mathbb{N}\rbrace$

Demonstration: A

It $(s_n)$ a non-decreasing monotonous succession. If the succession converges then it is bound (above); this demonstrates an implication of section A

Now suppose that the succession is superiorly bound, be it to its supreme and we will see that the succession converges to the point a

It $\epsilon > 0$ and as $a-\epsilon < a$, the number $a-\epsilon$ it may not be an upper bound of the sequence and therefore there will be some $N\in\mathbb{N}$ such that $a-\epsilon < s_N$

As succession is non-diminishing, for every $n > N$ you have to $a-\epsilon < s_N \leq s_{N+1}\leq \cdots\leq s_n$ and therefore, $a-\epsilon < s_n < a + \epsilon$

Conclusion: the succession converges to the point

Demonstration: B

It $(s_n)$ an unsrising monotonous succession. If the succession converges then it is bound (lower); this demonstrates an implication of paragraph B

Now suppose the succession is lower, be b its limit and we will see that the succession converges to point b

It $\epsilon > 0$ and as $b-\epsilon > b$, the number $b-\epsilon$ may not be a lower bound of the sequence and therefore there will be some $N\in\mathbb{N}$ such that $b-\epsilon > s_N$

As succession is non-increasing, for every $n < N$ you have to $b-\epsilon > s_N \geq s_{N+1}\geq \cdots\geq s_n$ and therefore, $b-\epsilon > s_n > b + \epsilon$

Conclusion: the succession converges to point b

Proposition: Converging succession equivalences

Are $(s_n)$ and $(t_n)$ succession are convergent with limits $a = \underset {n \to \infty} {\lim} s_n$, $b = \underset {n \to \infty} {\lim} t_n$ and $c \in \mathbb{C}$

• $(s_n + t_n)$ is convergent and has limit $a + b$
• $(c \cdot s_n)$ is convergent and has limit $c \cdot a$
• $(s_n \cdot t_n)$ is convergent and has limit $a \cdot b$
• If the succession $(t_n)$ contains no terms null and $b \not= 0$ then $(\frac{s_n}{t_n})$ is convergent and has limit $(\frac{a}{b})$

Demonstration: A

It $\epsilon > 0$
$a = \underset {n \to \infty} {\lim} s_n \Rightarrow \exists \omega_1$ such that $\left\|s_n - a\right\| < \frac{\epsilon}{2}; \forall n > N_1$ with $n, N_1 \in \mathbb{N}$
$b = \underset {n \to \infty} {\lim} t_n \Rightarrow \exists \omega_2$ such that $\left\|t_n - b\right\| < \frac{\epsilon}{2}; \forall n > N_2$ with $n, N_2 \in \mathbb{N}$
Then $N > \max\lbrace N_1, N_2 \rbrace$, therefore if $n > N$ is true that $\left\|(s_n + t_n) - (a + b)\right\| \leq \left\|(s_n - a)\right\| + \left\|(t_n - b)\right\|$ < $\frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Conclusion: the succession $(s_n + t_n)$ is convergent and has limit $a + b$

Demonstration: B

If $c = 0$ the result is trivial. So we'll assume that $c \not= 0$
$\exists n, N_1 \in \mathbb{N}$ such that $n > N_1$ then $\left\|s_n - a\right\| < \frac{\epsilon}{\left\|c\right\|}$
Then $\left\|c \cdot s_n - c \cdot a\right\| = \left\|c\right\| \cdot \left\|s_n - a\right\| < \frac{\left\|c\right\| \cdot c}{\left\|c\right\|} = \epsilon$

Conclusion: the succession $(c \cdot s_n)$ is convergent and has limit $c \cdot a$

Demonstration: C

It $\epsilon > 0$
$a = \underset {n \to \infty} {\lim} s_n \Rightarrow \exists \omega_1$ such that $\left\|s_n - a\right\| < \frac{\epsilon}{2 \cdot \left\|b\right\| + 1}; \forall n > N_1$ with $n, N_1 \in \mathbb{N}$
$b = \underset {n \to \infty} {\lim} t_n \Rightarrow \exists \omega_2$ such that $\left\|t_n - b\right\| < \frac{\epsilon}{2 \cdot K}; \forall n > N_2$ with $n, N_2 \in \mathbb{N}$
Then $\left\|s_n \cdot t_n - a \cdot b\right\| = \left\|s_n \cdot t_n - s_n \cdot b - a \cdot b \right\| \leq \left\|s_n\right\| \cdot \left\|t_n - b\right\| + \left\|b\right\| \cdot \left\|s_n - a\right\| \leq K \cdot \frac{\epsilon}{2 \cdot K} + \left\|b\right\| \cdot \frac{\epsilon}{2 \cdot \left\|b\right\| + 1} < \epsilon$

Conclusion: the succession $(s_n \cdot t_n)$ is convergent and has limit $a \cdot b$

Demonstration: D

It $\epsilon > 0$
$\left\|\frac{s_n}{t_n}-\frac{a}{b}\right\| = \left\|\frac{s_n \cdot b - t_n \cdot a}{b \cdot t_n}\right\| = \left\|\frac{s_n \cdot b - t_n \cdot a}{\left\|b\right\| \cdot \left\|t_n\right\|}\right\| = \left\|\frac{s_n \cdot b + s_n \cdot t_n - s_n \cdot t_n - t_n \cdot a}{\left\|b\right\| \cdot \left\|t_n\right\|}\right\| \leq \left\|\frac{\left\|s_n\right\| \cdot \left\|t_n - b\right\| + \left\|t_n\right\| \cdot \left\|s_n - a\right\|}{\left\|b\right\| \cdot \left\|t_n\right\|}\right\|$

Then $\exists n, N_1\in \mathbb{N}$ such that if $n > N_1$ you have to $\left\|t_n\right\| > \frac{\left\|b\right\|}{2}$

By be $s_n$ and $t_n$ inheritance of convergent $\exists K_1, K_2 > 0; \forall n \in \mathbb{N}$ such that $\left\|s_n\right\| < K_1$ and $\left\|t_n\right\| < K_2$

Then for the limit $s_n$ $\exists n, N_2\in \mathbb{N}$ such that if $n > N_2$ you have to $\left\|s_n - a\right\| < \frac{\epsilon \cdot \left\|b\right\|^{2} }{4 \cdot K_2}$

Then for the limit $t_n$ $\exists n, N_3\in \mathbb{N}$ such that if $n > N_3$ you have to $\left\|t_n - b\right\| < \frac{\epsilon \cdot \left\|b\right\|^{2} }{4 \cdot K_1}$

Then $n > \max\lbrace N_1, N_2, N_3 \rbrace$, therefore if $n > N$ is true that $\left\|\frac{s_n}{t_n}-\frac{a}{b}\right\| \leq \left\|\frac{\left\|s_n\right\| \cdot \left\|t_n - b\right\| + \left\|t_n\right\| \cdot \left\|s_n - a\right\|}{\left\|b\right\| \cdot \left\|t_n\right\|}\right\| < \frac{k_1 \cdot \frac{\epsilon \cdot \left\|b\right\|^{2} }{4 \cdot K_1} + k_2 \cdot \frac{\epsilon \cdot \left\|b\right\|^{2} }{4 \cdot K_2} }{\frac{\left\|b\right\| \cdot \left\|b\right\|}{2} } = \epsilon$

Conclusion: the succession $(\frac{s_n}{t_n})$ is convergent and has limit $(\frac{a}{b})$

Proposition: existence of limited intervals in convergent successions

Are $(s_n)$ and $(t_n)$ convergent successions and $\exists m | s_n \leqslant t_n \forall n > m$

Then $\underset {n \to \infty} {\lim} s_n \leqslant \underset {n \to \infty} {\lim} t_n$

Demonstration

The succession $t_n - s_n$ meets inequality $0 \leqslant t_n - s_n, \forall n > m$ and converges to $\underset {n \to \infty} {\lim} t_n - \underset {n \to \infty} {\lim} s_n$

We replace inequality $0 \leqslant \underset {n \to \infty} {\lim} t_n - \underset {n \to \infty} {\lim} s_n$, therefore it is fulfilled that $\underset {n \to \infty} {\lim} s_n \leqslant \underset {n \to \infty} {\lim} t_n$

Conclusion: the bounded interval exists $s_n \leqslant t_n$ for convergent successions $(s_n)$ and $(t_n)$

Cantor's Theorem of embedded intervals

As a result of the proposition of limited intervals in convergent successions we can list the Theorem of Cantor of the fitted intervals so that the intersection is formed by a single point

$\forall n \in \mathbb{N}$, it $I_n =[ a_n, b_n] \leq \emptyset$ a closed interval

Suppose that $I_{n+1}\subseteq I_n$ that is to say $a_n\leqslant a_{n+1}\leqslant b_{n+1}\leqslant b_n$ and that in addition $\underset {n \to \infty} {\lim} \left ( b_n - a_n \right ) = 0$

Then $\underset {\in \mathbb{N}} {\bigcap} I_n = \left \{ x \right \}$ where $x = \underset {n \to \infty} {\lim} a_n = \underset {n \to \infty} {\lim} b_n$

Demonstration

By hypothesis, succession $(a_n)$ is monotonous, non-decreasing and quoted superiorly (for example $b_1$), therefore converging $x\in\mathbb{R}$

Similarly, succession $(b_n)$ converges to $r\in\mathbb{R}$ and by the previous proposition $a_n\leqslant x \leqslant r \leqslant b_n, \forall n \in\mathbb{N}$

We replace the condition $\underset {n \to \infty} {\lim} (b_n - a_n) = 0$ that assures us that $x = r$ and that $\left \{ x \right \} = \underset {n \in \mathbb{N}} {\bigcap} I_n$

With what is proven the Cantor's Theorem of embedded intervals

Proposition: Sandwich Rule

Are $(s_n)$, $(t_n)$ and $(u_n)$ and $\exists m \in\mathbb{N} | s_n \leqslant t_n \leqslant u_n, \forall n > m$

If $(s_n)$ and $(u_n)$ are convergent successions and $\underset {n \to \infty} {\lim} s_n = \underset {n \to \infty} {\lim} u_n = a$

Then $(t_n)$ is also convergent and $\underset {n \to \infty} {\lim} t_n = a$

Demonstration

It $\epsilon > 0$, by the definition of limit $\exists N_1 \in \mathbb{N}\Rightarrow \text{if }n > N_1 \rightarrow \left \| s_n - a \right \| < \epsilon$

That is to say $a - \epsilon < s_n < a + \epsilon$

Similarly $\exists N_2 \in \mathbb{N}\Rightarrow \text{if }n > N_2 \rightarrow \left \| u_n - a \right \| < \epsilon$

So if $n > \text{m\'{a}x}\left \{ m, N_1, N_2 \right \}$ you have to $a - \epsilon < s_n \leqslant t_n \leqslant u_n < a + \epsilon$ or what's equivalent $\left \| t_n -a \right \| < \epsilon$

With what we've tried the Sandwich Rule