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Given $x_n$ is convergent sequence with limit eqals to $a$. to prove that

$y_n = \frac{x_1 + x_2 + ... + x_n}{n}$ converges to same limit

Now $$\left|\frac{x_1 + x_2 + ... + x_n}{n} - a \right|\\=\left|\frac{(x_1 - a) + (x_2-a) + ... + (x_n-a)}{n}\right |$$

Since $x_n$ is convergent. so it is bounded. so for all $n$

$x_n \leq M + a$

I am not sure how to proceed. Thanks

Thomas Andrews
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Olivia
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    https://math.stackexchange.com/q/155839/42969, https://math.stackexchange.com/q/2858957/42969, https://math.stackexchange.com/q/193157/42969 – Martin R Mar 23 '23 at 14:55
  • Look up “cesaro mean convergence proof” – this has been asked and answered many times. – Martin R Mar 23 '23 at 14:58
  • @MartinRi am confused if numerator has n terms or infinite terms..i mean is $x_{n+1}$ part of numerator ? – Olivia Mar 23 '23 at 15:00
  • Every numerator has finitely many terms: $y_1 = x_1$, $y_2 = (x_1+x_2)/2$, $y_3 = (x_1 + x_2 + x_3)/3$, ... – Martin R Mar 23 '23 at 15:03
  • @MartinR but $y_n$ are infinite, no ? – Olivia Mar 23 '23 at 15:03
  • Of course there are infinitely many $y_n$, but each $y_n$ is the sum of finitely many $x_n$. Where is the problem? – Martin R Mar 23 '23 at 15:06
  • Your equality shows we really only need to show if $x_n\to0,$ then the means converge to $0.$ – Thomas Andrews Mar 23 '23 at 15:11
  • So when $x_n\to0,$ we have that all $|x_i|<M$ for some $M,$ and for $i$ large, $|x_i|$ is small. So you need "enough" of the small terms to outweigh the terms we know are only bounded. So, find $N$ such that $|x_i|<\epsilon/2$ for $i>N.$ Then for $n>N$ we have $$|y_n|\leq \frac{NM}{n}+\frac\epsilon 2\frac{n-N}{n}\leq \frac\epsilon2+\frac{MN}n.$$ So you need $$\frac{NM}{n}<\frac\epsilon2$$ or $$n>\frac{2MN}{\epsilon}.$$ Set $N'=\max(N,2MN/\epsilon)$ and you are done. – Thomas Andrews Mar 23 '23 at 15:21

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Your equality shows that we can restrict to the case when $x_n\to0,$ since, if $x_n\to a,$ we can solve the problem for $z_n=x_n-a\to0.$

So when $x_n\to0,$ we have that there exists an $M$ such that all $|x_i|<M,$ and for $i$ large, $|x_i|$ is small.

So you need "enough" of the small terms to outweigh the terms we know are only bounded.

So, find $N$ such that $|x_i|<\epsilon/2$ for $i>N.$ Then for $n>N$ we have $$\begin{align}|y_n|&\leq \frac1n\sum_{i=1}^n|x_i|\\ &=\frac1n\sum_{i=1}^N|x_i|+\frac1n\sum_{i=N+1}^n|x_i|\\ &\leq\frac{NM}{n}+\frac\epsilon 2\frac{n-N}{n}\\ &\leq \frac\epsilon2+\frac{MN}n.\end{align} $$

So you need $$\frac{NM}{n}<\frac\epsilon2$$ or $$n>\frac{2MN}{\epsilon}.$$

Set $N'=\max(N,2MN/\epsilon)$ and then for $n>N',$ $|y_n|<\epsilon.$

Essentially, $N$ determines that the first $N$ terms are merely known to be bounded, and any terms after these are "small." Then $MN/\epsilon$ defines what it means for "enough terms to be small" for the small terms to outweigh those first $N$ terms.

Thomas Andrews
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  • This has been asked and answered many times: https://math.stackexchange.com/questions/linked/155839. Is it really necessary to post another proof? – Martin R Mar 23 '23 at 15:50
  • I hated to see the wrong answer as the only answer. @MartinR Initially, I was content to put it in comments. – Thomas Andrews Mar 23 '23 at 15:51
  • But made my answer community wiki to ensure I'm not just gaming the system. – Thomas Andrews Mar 23 '23 at 15:53
  • maybe $|x_i| < \frac {\epsilon} { 2} $ for $i \geq n+1$...so what to do then – Olivia Mar 23 '23 at 15:59
  • Don't confusing $n$ and $N.$ Different variables. I have written more after that step, what didn't you follow? Be specific. @Olivia – Thomas Andrews Mar 23 '23 at 16:01
  • @ThomasAndrews why you split the n terms into two pieces (is it because boundedness alone wont solve it ?) ? and how to split them if the $x_i$ is less than epsilon for very large value which cross n. like example sequence x_n lies in epsilon neighbourhood for value more than n, say n+9 and is the little n the original n which is to be found for given epsilon ? – Olivia Mar 23 '23 at 16:06
  • Specifically, $N$ is a fixed single value, depending on the sequence $(x_i)_{i=1}^{\infty}$ and $\epsilon.$ – Thomas Andrews Mar 23 '23 at 16:13
  • yes but y_n has n terms of x_n only and not infinite. – Olivia Mar 23 '23 at 16:16
  • As I said at the beginning and end, you need enough small terms to outweigh the terms we only know are bounded. So you want to split the terms into "known to be small" and "not-known to be small." Luckily, $x_n\to0$ means we can make a cut-off that lets us say "every term after this is small." – Thomas Andrews Mar 23 '23 at 16:16
  • Who said anything about infinite terms? @Olivia – Thomas Andrews Mar 23 '23 at 16:16
  • but all terms are bounded anyhow i.e smaller than something. so what is known to be small and not known to be small – Olivia Mar 23 '23 at 16:18
  • if we assume this at start $x_i < \epsilon $ for $n> n+1, n+2,n+3... $ – Olivia Mar 23 '23 at 16:19
  • $n>n+1?$ When is it possible for $n>n+1?$ $n>N.$ Don't mistake the $n$ and $N.$ @Olivia I have no idea what you are getting at in that comment. – Thomas Andrews Mar 23 '23 at 16:25
  • You are correct, being bounded is not enough. The whole point is that being bounded is not enough, that we can cut off and consider some finite number of $x_i$ as potentially problematic, and make the number of terms which aren't problematic large enough to outweigh the problematic ones. @Olivia – Thomas Andrews Mar 23 '23 at 16:29
  • An example would be $x_k=\frac1k$ and $\epsilon=\frac14.$ Then we can choose $M=1, N=8.$ and we conclude that if $n>2MN/epsilon=64,$ then $$|y_n|=\frac{1+\frac12+\cdots+\frac1n}{n}<\frac14.$$ – Thomas Andrews Mar 23 '23 at 16:30
  • @ThomasAndrews one thing i was doubt is that if tail of x_n starts after i > n then what to do ? because we have assumed that tail starts from somewhere at N+1. – Olivia Mar 24 '23 at 14:23
  • You've confused $n$ with $N,$ again. There is an $N$ such that $|x_i|<\epsilon/2$ for $i>N,$ by the definition of $\lim_{n\to\infty}x_n=0.$ If we don't know $x_n\to 0,$ we don't have the rest of the proof. Maybe you are confused by the phrase $x_n\to0?$ It is just a shorthand for $\lim_{n\to\infty}x_n=0.$ – Thomas Andrews Mar 24 '23 at 16:22