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Suppose that $\{a_n\}$ is a real sequence with $$\lim_{n\to\infty}\frac{\sum\limits_{k=1}^na_k}{n}=0,\lim_{n\to\infty}(a_{n+1}-a_n)=0,$$ then can we get $$\lim_{n\to\infty}a_n=0?$$

This simple problem has got on my nerves for two days, I've tried to prove that is ture, however, there's nothing I can get.

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    $\begingroup$ Perhaps you should learn about Tauberian theorems... $\endgroup$
    – GEdgar
    Dec 30, 2013 at 14:41
  • $\begingroup$ @GEdgar , can you give an explicit direction? $\endgroup$
    – Gamezeta
    Dec 30, 2013 at 15:03
  • $\begingroup$ Instead of $\lim\limits_{n→∞}(a_{n+1}−a_n)=0$, you should use $\lim\limits_{n→∞}n(a_{n+1}−a_n)=0$. Then the expression will be true. $\endgroup$
    – guest
    Oct 10, 2014 at 9:01

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The proposed theorem seems to be false. I think that the series $a_k = \sin( \sqrt{k})$ is a counterexample.

The difference between $a_{n+1}$ and $a_n$ is at most $|\sqrt{n+1} -\sqrt{n}|$ because the derivative of sin is between -1 and 1, so it tends to zero.

It's clear that the series $a_n$ does not tend to zero, since its values approach 1 and -1 over and over forever. It remains to show that the mean of $\{a_k\}$ tends to zero. This technical feat is beyond me at the moment, I'm afraid.

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  • $\begingroup$ If you modify this to $\sin(\pi\sqrt{k})$, then there is a full wave between every $k=n^2$ and $k=(n+2)^2$. Modify further to have the arguments in that segment equally spaced, then the average should converge to zero. Or alternatively, based on $\sin(\pi\log_2(n)$, $a_{2^n+k}=\sin(k\,2^{1-n}\pi)$ for $k=0,1,\dots,2^{n}-1$. $\endgroup$
    – Lutz Lehmann
    Dec 30, 2013 at 17:31
  • $\begingroup$ I had come up with the way to cope with the limit that $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\sin\sqrt k=0.$$ $\endgroup$
    – Gamezeta
    Jan 1, 2014 at 4:00
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    $\begingroup$ Lemma. If the series $\displaystyle\sum_{n=1}^\infty\frac{a_n}{n}$ is convergent, then we have$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^na_k=0.$$ $\endgroup$
    – Gamezeta
    Jan 1, 2014 at 4:04
  • $\begingroup$ Lemma. Suppose that $f\in C^1[1,+\infty)$ with $$\int_1^\infty|f'(x)|dx<+\infty,$$then the integration $\displaystyle\int_1^\infty f(x)dx$ has the same convergence as the series $\displaystyle\sum_{n=1}^\infty f(n).$ Anyway, thanks for your counterexample. $\endgroup$
    – Gamezeta
    Jan 1, 2014 at 4:08

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