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Given $a_n$ be a sequence and IF $\lim_{n\to\infty}a_{n}=l$, Then prove that $\lim_{n\to\infty}\frac{a_{1}+a_2+\cdot..+a_n}{n}=l$

I do not know how to do this. Can someone help me with this?

Thanks

ATTEMPT

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marked as duplicate by Martin Sleziak, Alexis Olson, Claude Leibovici, tired, Pragabhava Oct 18 '16 at 19:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Welcome to Stackexchange. You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult "How to ask a good question?" for further guidance. $\endgroup$ – Matthew Leingang Aug 18 '15 at 17:13
  • $\begingroup$ It's consequence from Stolz-Cesaro theorem $\endgroup$ – Michael Galuza Aug 18 '15 at 17:16
  • $\begingroup$ See this and its "linked' section. $\endgroup$ – David Mitra Aug 18 '15 at 17:17
  • $\begingroup$ @MatthewLeingang "Welcome to Math Stack"...the Op has 739 rep. If you're going to copy and paste statements at least see if they apply... $\endgroup$ – Zach466920 Aug 18 '15 at 17:18
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    $\begingroup$ @Zach466920. Good point. It's too late for me to edit that part of the comment, but in the future I'll keep that in mind. $\endgroup$ – Matthew Leingang Aug 18 '15 at 17:23
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Hint:

Given $\epsilon > 0$, find $N$ so that $|a_k-L|<\epsilon/2$ for $k > N$. Then

$$\left | \frac{\sum_{k=1}^n a_k}{n} - L \right | = \left | \frac{\sum_{k=1}^n a_k - L}{n} \right | \leq \frac{\sum_{k=1}^n |a_k-L|}{n}.$$

By how this was already set up, the contribution to the average from the terms with $k > N$ is already less than $\epsilon/2$. Now try to take $n$ a bit larger, so that the contribution to the average from the terms with $k \leq N$ is also less than $\epsilon/2$.

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  • $\begingroup$ Please check my proof, i think i was on same lines $\endgroup$ – Taylor Ted Aug 18 '15 at 17:55
  • $\begingroup$ @TaylorTed Followup hint: prove that $|a_k-L| \leq M$ for some $M$. Then replace the terms with low index by this $M$ and see what you get. $\endgroup$ – Ian Aug 18 '15 at 17:56
  • $\begingroup$ I did not quite follow you. Can we go step by step please $\endgroup$ – Taylor Ted Aug 18 '15 at 18:01
  • $\begingroup$ @TaylorTed A convergent sequence is bounded, say by $M$. By the triangle inequality, $|a_k-L| \leq M+|L|$. Call this $M'$. Now split the sum up, into $\frac{\sum_{k=1}^{N} |a_k-L|}{n}+\frac{\sum_{k=N+1}^n |a_k-L|}{n}$, and replace all of the $|a_k-L|$ in the first term with $M'$. Now what happens? $\endgroup$ – Ian Aug 18 '15 at 18:07
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Let $x_n=a_1+a_2+...+a_n$ and $y_n=n$. Now, $\displaystyle \lim_{n\rightarrow \infty}\frac {x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\rightarrow \infty}a_{n+1}=\lim_{n\rightarrow \infty}a_{n}=l$. So by Cesaro-Stolz’s theorem, $\displaystyle \lim_{n\rightarrow \infty}\frac{x_n}{y_n}=\lim_{n\rightarrow \infty}\frac{a_1+a_2+...+a_n}{n}=l.$

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You could just apply Stolz-Cesaro blindly or to gain some insight make an argument along the following lines.

Since $a_n \to l$, for any $\epsilon>0$ there is a positive integer $N$ such that $l- \epsilon < a_n < l + \epsilon$ when $n > N$.

Hence, with $S_n = a_1 + a_2 + \ldots + a_n$,

$$\frac{1}{n}S_n = \frac1{n}\sum_{k=1}^{N}a_k + \frac1{n}\sum_{k=N+1}^{n }a_k \\\leqslant \frac1{n}\sum_{k=1}^{N}a_k + \frac1{n}(n-N)(l + \epsilon),$$

and

$$\limsup_{n \to \infty}\frac{1}{n}S_n \leqslant l + \epsilon.$$

Make a similar argument to show that

$$\liminf_{n \to \infty}\frac{1}{n}S_n \geqslant l - \epsilon.$$

Since both inequalities hold for any $\epsilon> 0$ we have

$$\limsup_{n \to \infty}\frac{1}{n}S_n = \liminf_{n \to \infty}\frac{1}{n}S_n = \lim_{n \to \infty}\frac{1}{n}S_n=l$$

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  • $\begingroup$ @Nitin As a fellow UC alum, I am a bit surprised as to your dismissal of CS. $\endgroup$ – Mark Viola Aug 18 '15 at 17:39
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    $\begingroup$ @Dr.MV indeed, proving this statement without CS on an analysis exam last year was a pain, but using a jackhammer to kill a fly is always fun to watch. $\endgroup$ – user217285 Aug 18 '15 at 17:42
  • $\begingroup$ @Nitin Now that is funny! +1 just because. $\endgroup$ – Mark Viola Aug 18 '15 at 17:44
  • $\begingroup$ @Dr.MV: $|a_k - l| < \epsilon \implies -\epsilon < a_k - l < \epsilon$. I did not use the triangle inequality. $\endgroup$ – RRL Aug 18 '15 at 17:46
  • $\begingroup$ @RRL Yes, I see. I missed the second part in which you apply the left side inequality. Well done then. +1 $\endgroup$ – Mark Viola Aug 18 '15 at 17:51
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Note that

$$\lim_{n\to 0}a_n=\lim_{n\to 0}\frac{\sum_{k=1}^na_k-\sum_{k=1}^{n-1}a_k}{(n)-(n-1)}=\ell$$

By the Stolz-Cesaro Theorem we have immediately that

$$\lim_{n\to 0}\frac{\sum_{k=1}^na_k}{n}=\ell$$

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 28 '15 at 19:29

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