Convergence

We are going to deal with the convergence of sequences deduced linearly from others based on the book by Julio Rey Pastor, Algebraic Analysis

General Criterion

Theorem

It (a_n) with n\geq 1 a convergent limit sequence \alpha and we consider the coefficients (\lambda_{i,n}) with n\geq 1, 1\leqslant i\leqslant n that verify:

• \underset {n \to \infty} {\lim} \lambda_{i,n} = 0, \forall i
• \underset {n \to \infty} {\lim} \lambda_{1,n} + \lambda_{2,n} + \cdots + \lambda_{n,n} = \lambda
• And if the \lambda_{i,j} they are not positive, in addition, \left\|\lambda_{1,n}\right\| + \left\|\lambda_{2,n}\right\| + \cdots + \left\|\lambda_{n,n}\right\| \leqslant K

Demonstration

Given that c_n=a_1 \cdot \lambda_{1,n} + a_2 \cdot \lambda_{2,n} + \cdots + a_n \cdot \lambda_{n,n} = \alpha \cdot (\lambda_{1,n} + \lambda_{2,n} + \cdots + \lambda_{n,n}) + (a_1 - \alpha) \cdot \lambda_{1,n} + (a_2 - \alpha) \cdot \lambda_{2,n} + \cdots + (a_n - \alpha) \cdot \lambda_{n,n}

If we use \underset {n \to \infty} {\lim} a_n = \alpha , we have to given \epsilon > 0, \exists n_0 such that \left\|a_n - \alpha \right\| < \frac{\epsilon}{3\cdot K}, \forall n \geqslant n_0 with K the value of property C)

For property A) there will be a value n_1, so we will take a value greater than n_0 such that \left\|\lambda_{i,n}\right\| < \frac{\epsilon }{3 \cdot (\left\|a_1 - \alpha \right\| + \cdots + \left\|a_{n_0 - 1} - \alpha \right\|) }, n\geqslant n_1 and 1 \leqslant i < n_0

By property B) we can deduce that there will be n_2, so we will take a value greater than n_1 such that \left\|\lambda_{1,n} + \lambda_{2,n} + \cdots + \lambda_{n,n} - \lambda \right\| < \frac{\epsilon }{3 \cdot \left\| \alpha \right\|}, \forall n \geq n_2

Using the previous results and the triangular inequality we have:

\left\| c_n - \alpha \cdot \lambda \right\| \leqslant \left\| \alpha \right\| \cdot \left\| \lambda_{1,n} + \lambda_{2,n} + \cdots + \lambda_{n,n} - \lambda \right\| + \left\| a_1 - \alpha \right\| \cdot \left\| \lambda_{1,n} \right\| + \cdots + \left\| a_{n_0 - 1} - \alpha \right\| \cdot \left\| a{{n_0 - 1},n} \right\| + \left\| a_{n_0} - \alpha \right\| \cdot \left\| a_{{n_0},n} \right\| + \cdots + \left\| a_n - \alpha \right\| \cdot \left\| \lambda_{n,n} \right\| < \left\| \alpha \right\| \cdot \frac{\epsilon}{3 \cdot \left\| \alpha \right\|} + \frac{\epsilon \cdot (\left\| a_1 - \alpha \right\| + \cdots + \left\| a_{n_0-1} - \alpha \right\|) }{3 \cdot (\left\| a_1 - \alpha \right\| + \cdots + \left\| a_{n_0-1} - \alpha \right\|)} + \frac{\epsilon}{3\cdot K} \cdot(\left\| \lambda_{n_0, n} \right\| + \cdots + \left\| \lambda_{n,n} \right\|) \leq \epsilon

Applying property C) in the last step, with which we have tested the general criterion

Notes on the general criterion

Let us consider a matrix with infinite rows and columns, in such a way that in the nth row we place the values (\lambda_{i, n}) with 1 \leq i \leq n and we complete the row with 0
A=\begin{pmatrix} \lambda_{1,1} & 0 & 0 & 0 & \cdots & 0 & 0 & \cdots \\ \lambda_{1,2} & \lambda_{2,2} & 0 & 0 & \cdots & 0 & 0 & \cdots \\ \lambda_{1,3} & \lambda_{2,3} & \lambda_{3,3} & 0 & \cdots & 0 & 0 & \cdots \\ \lambda_{1,4} & \lambda_{2,4} & \lambda_{3,4} & \lambda_{4,4} & \cdots & 0 & 0 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots &\cdots & \cdots \\ \lambda_{1,n} & \lambda_{2,n} & \lambda_{3,n} & \lambda_{4,n} & \cdots & \lambda_{n,n} & 0 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots &\cdots & \cdots \end{pmatrix}

With this matrix A, the sequence c_n defined in the previous proof is obtained as:

\begin{pmatrix} c_1\\ c_2\\ c_3\\ c_4\\ \cdots\\ c_n\\ \cdots \end{pmatrix}= A \cdot \begin{pmatrix} a_1\\ a_2\\ a_3\\ a_4\\ \cdots\\ a_n\\ \cdots \end{pmatrix}=\begin{pmatrix} \lambda_{1,1} & 0 & 0 & 0 & \cdots & 0 & 0 & \cdots \\ \lambda_{1,2} & \lambda_{2,2} & 0 & 0 & \cdots & 0 & 0 & \cdots \\ \lambda_{1,3} & \lambda_{2,3} & \lambda_{3,3} & 0 & \cdots & 0 & 0 & \cdots \\ \lambda_{1,4} & \lambda_{2,4} & \lambda_{3,4} & \lambda_{4,4} & \cdots & 0 & 0 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots &\cdots & \cdots \\ \lambda_{1,n} & \lambda_{2,n} & \lambda_{3,n} & \lambda_{4,n} & \cdots & \lambda_{n,n} & 0 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots &\cdots & \cdots \end{pmatrix}\cdot \begin{pmatrix} a_1\\ a_2\\ a_3\\ a_4\\ \cdots\\ a_n\\ \cdots \end{pmatrix}

It is interesting that property A) of the theorem is telling us that each column of matrix A tends to 0 when n \to \infty

Property B) tells us that if we add the elements of each row and calculate the limit of those sums, when n \to \infty, the limit is \lambda

Consequences of the general criterion

Properly applying the general criterion we can deduce some interesting results, some of them already known and others totally new

Limit of the arithmetic mean and the geometric mean of a sequence

Let's take \lambda_{i,n}=\frac{1}{n} with i=1, \cdots, n

Whose positive coefficients satisfy A) and \lambda_{1,n} + \lambda_{2,n} + \cdots + \lambda_{n,n} = \frac{1}{n} + \cdots + \frac{1}{n} = 1, then satisfy B) when \lambda = 1

If we consider a sequence (a_n)_{n\geq 1} with limit \alpha, by the general criterion it will fulfill that:

\underset {n \to \infty} {\lim} \frac{a_1 + \cdots + a_n}{n} = \alpha

That is, if the sequence has a limit, the sequence of the arithmetic means of the first n terms have the same limit (Rey Pastor attributes this result to Cauchy)

Furthermore, if a_n > 0, taking logarithms, it will satisfy that:

\underset {n \to \infty} {\lim} \sqrt[n]{a_1 + \cdots + a_n} = \exp (\underset {n \to \infty} {\lim} \frac{\log a_1 + \cdots + \log a_n}{n}) = \exp (\log \alpha ) = \alpha

If we consider a positive sequence with limit, the sequence of geometric means of the first n terms have the same limit

Finally, given a sequence (b_n)_{n\geq 1}, if we apply the criterion of geometric means taking a_1 = b_1 and a_n = \frac{b_n}{b_{n-1}} (in this specific case we have to \sqrt[n]{a_1 + \cdots + a_n} = \sqrt[n]{b_n}), will fulfill that:

\underset {n \to \infty} {\lim} \frac{b_n}{b_{n-1}} = b \Rightarrow \underset {n \to \infty} {\lim} \sqrt[n]{b_n} = b

Stolz criterion

The Stolz criterion is a particular case of convergence of a series deduced linearly from another

It (b_n)_{n\geqslant 1} a succession of positive terms such that B_n = b_1 + \cdots + b_n\to \infty, n \to \infty and (a_n)_{n\geqslant 1} a convergent sequence with limit \alpha, will fulfill that:

\underset {n \to \infty} {\lim} \frac{a_1 \cdot b_1 + \cdots + a_n \cdot b_n}{B_n} = \alpha

Following the general criterion we take \lambda_{i,n} = \frac{b_i}{B_n}, for i = 1, \cdots, n, will fulfill that:

\underset {n \to \infty} {\lim} \frac{b_i}{B_n} = 0

So A) of the general criterion is met and \lambda_{1,n} + \lambda_{2,n} + \cdots + \lambda_{n,n} = \frac{b_1 + \cdots + b_n}{B_n} = 1, so B is also true)

In particular, given a sequence (d_n)_{n\geqslant 1} and (b_n)_{n\geqslant 1}, taking the previous result a_n = \frac{d_n}{b_n}, it follows that:

\underset {n \to \infty} {\lim}\frac{d_n}{b_n} = l \Longrightarrow \underset {n \to \infty} {\lim}\frac{d_1 + \cdots + d_n}{b_1 + \cdots + b_n} = l

And if we take a succession (B_n)_{n\geqslant 1} divergent, shall comply with the fact that:

\underset {n \to \infty} {\lim}\frac{D_1 + \cdots + D_n}{B_1 + \cdots + B_n} = l \Longrightarrow \underset {n \to \infty} {\lim}\frac{D_n}{B_n} = l

It will be enough for us to take d_n = D_n - D_{n-1} for the previous case

This result is known as the Strolz criterion

Consequence I

Are (a_n)_n{n\geqslant 1} and (b_n)_n{n\geqslant 1} two sequences such that A_n = a_1 + \cdots + a_n and B_n = b_1 + \cdots + b_n

If \underset {n \to \infty} {\lim}A_n = A, \underset {n \to \infty} {\lim} B_n = B and \left\|b_1\right\| + \cdots + \left\|b_n\right\| \leqslant K, will fulfill that:

\underset {n \to \infty} {\lim} A_1 \cdot b_n + A_2 \cdot b_{n-1} + \cdots + A_n \cdot b_1 = A \cdot B

This result is a consequence of the general criterion taking \lambda_{i,n} = b_{n-i+1} with i = 1, \cdots, n

Taking into account that \underset {n \to \infty} {\lim}b_{n-i+1} =\underset {n \to \infty}B_{n-i+1} - B_{n-i} = B - B = 0, which implies A)

Of the condition \underset {n \to \infty} {\lim} B_n = B B) and C) are deduced (in this case we should have checked that they were positive, so we use the condition \left\|b_1\right\| + \cdots + \left\|b_n\right\| \leqslant K)

Consecuencia II

Are (a_n)_n{n\geqslant 1} and (b_n)_n{n\geqslant 1} two sequences such that \underset {n \to \infty} {\lim}a_n = \alpha and \underset {n \to \infty} {\lim}b_n = \beta, will fulfill that:

\underset {n \to \infty} {\lim} a_1 \cdot b_n + a_2 \cdot b_{n-1} + \cdots + a_n \cdot b_1 = \alpha \cdot \beta

This result is a consequence of the general criterion taking \lambda_{i,n} = \frac{b_{n-i+1}}{n}

To check A) we must observe the convergence of the sequence b_n that assures us that \left\|b_n\right\|\leqslant C, then \left\|\lambda_{i,n}\right\|\leqslant \frac{C}{n}\to 0, n\to\infty

If we apply the Stolz criterion we can check B) (we could also have applied the arithmetic means criterion):

\underset {n \to \infty} {\lim} \lambda_{1,n} + \lambda_{2,n} + \cdots +\lambda_{n,n} = \underset {n \to \infty} {\lim} \frac{b_1 + \cdots + b_n}{n} = \underset {n \to \infty} {\lim} b_n = \beta

Condition C) is again a consequence of the constraint of the sequence b_n since it is fulfilled that \left\| \lambda_{1,n} \right\| + \left\| \lambda_{2,n} \right\| + \cdots + \left\| \lambda_{n,n} \right\| \leqslant C \cdot (\frac{1}{n} + \cdots + \frac{1}{n}) = C