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Theorem.

If $\sum_{n=1}^{\infty}c_{n}$ is Cesaro summable, then $c_{n}/n$ tends to $0$.

How to prove it?

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    $\begingroup$ Your title does not correspond to your question. $\endgroup$ – Chris Eagle Oct 6 '12 at 13:33
  • $\begingroup$ Please consider edit your post according to what you want to ask $\endgroup$ – leo Oct 6 '12 at 15:19
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Hint: Cesàro summable means that $\lim_{n\to\infty}\frac{c_1+\cdots+c_n}n$ exists. Note that $\frac{c_1+\cdots+c_n}n = \frac{c_1+\cdots+c_{n-1}}{n-1}\cdot(1-\frac1n)+\frac{c_n}n$.

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  • $\begingroup$ Thank you for your advice! $\endgroup$ – Guillermo Oct 6 '12 at 13:36

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