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I have the following sequence :

$$b_n := \frac1n * \sum_{k=1}^n a_k$$

and I want to prove that this sequence converges. I know that the sequence $a_n$ converges already. I am not quite sure however where to start with the proof. For example: How do I prove that atleast the sum of a convergent sequence converges again? Maybe by using cauchy sequences? I am a bit lost so any hint/help would be kindly appreciated.

Jose M Serra
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Jack4t3
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  • The question has been posted numerous times here, so search for a duplicate. – DeepSea Nov 26 '16 at 10:52
  • you might start with witing $b_n=\sum_{k=1}^{n-1}b_{n+1}-b_n$ as series and then consider the summands $b_{n+1}-b_n$. i did not check weather this works, it might or might not help. – Max Nov 26 '16 at 10:52
  • i messed up the indices, sorry for that. the summands on the right hand side must of course be indiced by $k$ and $k+1$ instead of $n$ and $n+1$ – Max Nov 26 '16 at 11:02
  • A very similar question is here. The difference is that the linked question already assume that the Césaro sum will converge to the same limit. – Masacroso Nov 26 '16 at 11:49

2 Answers2

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This is known as Cauchy's First Theorem on Limits:

$\lim_{n\to \infty} a_n<\infty\implies \lim_{n\to \infty}\dfrac{a_1+a_2+\ldots +a_n}{n}<\infty$

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If $a_n\to0$ then for every $\varepsilon>0$ we can find an $M$ such that $\lvert a_k\rvert<\varepsilon$ for all $k>M$. This implies $$-\varepsilon\leftarrow\frac1n\sum_{k=1}^M a_k -\varepsilon<\frac1n\sum_{k=1}^M a_k -\frac{\varepsilon(n-M)}{n}=\frac1n\sum_{k=1}^M a_k -\frac{1}{n}\sum_{k=M+1}^n \varepsilon<b_n<\frac1n\sum_{k=1}^M a_k +\frac{1}{n}\sum_{k=M+1}^n \varepsilon= \frac1n\sum_{k=1}^M a_k +\frac{\varepsilon(n-M)}{n}<\varepsilon+\frac1n\sum_{k=1}^M a_k\to\varepsilon,$$and since $\varepsilon$ is arbitrary, the squeeze theorem yields $b_n\to0$.

If $a_n\to A\ne0$, it is eventually positive or negative. Hence $\sum_{k=1}^n a_k$ is eventually monotone increasing to $+\infty$ or monotone decreasing to $-\infty$. This means we can apply the Stolz-Cesàro theorem, which gives that $b_n$ has the same limit as $a_n$.

Vincenzo Oliva
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