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We have a well-known conclusion:

If a sequence $\{a_n\}_{n\in\mathbb{N}}$ converges, then the arithmetic mean $\frac{S_n}{n}$ (where $S_n=\sum\limits_{k=1}^na_k$ is the nth partial sum) converges to the same limit.

$\lim\limits_{n\to\infty}a_n=a\implies\lim\limits_{n\to\infty}\frac{S_n}{n}=a$

I hope the inverse process also works, so I add a condition:

Sequence $\{n(a_n-a_{n-1})\}_{n\in\mathbb{N}}$ is bounded, i.e,

$\exists M>0,\forall n\in\mathbb{N},|n(a_n-a_{n-1})|<M$

With such condition, can $\frac{S_n}{n}$ converges implies $a_n$ converges?

I want to prove that

$\lim\limits_{n\to\infty}\frac{S_n}{n}=a,\ |n(a_n-a_{n-1})|<M\implies \lim\limits_{n\to\infty}a_n=a$

I see that

$|a_n-\frac{S_n}{n}|=\frac{1}{n}|\sum\limits_{k=2}^n (k-1)(a_{k}-a_{k-1})| \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq\frac{1}{n}\sum\limits_{k=2}^n |(k-1)(a_{k} - a_{k-1})|\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq\frac{1}{n}\sum\limits_{k=2}^n |k(a_{k} - a_{k-1})|\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ < \frac{(n-1)M}{n}<M$

then $|a_n|\leq |\frac{S_n}{n}|+M$ means that $\{a_n\}$ is bounded.

Now I can't continue the proof (Maybe consider a convergent subsequence of$\{a_n\}$? I failed..). Does the added condition works? Could you prove that?

Sincerely looking forward to your help. Thanks!

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  • $\begingroup$ You need to look at Cesaro summability en.wikipedia.org/wiki/Cesàro_summation $\endgroup$ Aug 5, 2017 at 17:59
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    $\begingroup$ emmm...I can't understand how to use Cesaro-sum in this problem $\endgroup$
    – kellty
    Aug 5, 2017 at 18:08
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    $\begingroup$ You may want to have a look at this. $\endgroup$
    – user436658
    Aug 5, 2017 at 18:14
  • $\begingroup$ Thanks a lot! I'd turn to Rudin's book $\endgroup$
    – kellty
    Aug 5, 2017 at 18:24
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    $\begingroup$ Cesaro summability involves taking the average of the first terms to find a value - it is used in Fourier Analysis, for example, where (under useful conditions) it enables a useful value to be defined. Hardy's "Divergent Series" also covers this. The point is that the "averaging" technique sums more series than the conventional sum, and gives the same result as the conventional sum when that exists. $\endgroup$ Aug 5, 2017 at 19:06

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