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Suppose that $(a_n)$ is a sequence of real numbers such that series: $\sum_{n=1}^\infty \frac{a_n}{n}$ is convergent. Show that the sequence: $$b_n=\frac{ \sum_{j=1}^n a_j}{n}$$ is convergent and find its limit.

I try to solve it by Stolz theorem, but no success.

Source (http://vjimc.osu.cz/hist/j16problems1.pdf)

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1 Answer 1

Hint: Apply Cesàro to $$b_n=A_n-\frac1n\sum_{j=1}^{n-1}A_j,\qquad A_j=\sum_{k=1}^j\frac{a_k}k.$$

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Can anyone give more detail, please? –  user44636 Jun 18 at 10:38
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Sure--as soon as you explain what is blocking you. –  Did Jun 18 at 11:40
    
Following Did's hint one should rewrite $b_n$ as $\frac{1}{n} (nA_n - \sum_{j=1}^{n-1}A_j)$ and try to find limit of the sequence $(nA_n - \sum_{j=1}^{n-1}A_j)$, but I can't do it. –  user44636 Jun 18 at 12:40
    
No, one is not supposed to find the limit of $nA_n-\sum\limits_{j=1}^{n-1}A_j$ (what gave you this idea?), simply the limit of $A_n-\frac1n\sum\limits_{j=1}^{n-1}A_j$. That $A_n$ converges is a hypothesis. So what is left is to show that $\frac1n\sum\limits_{j=1}^{n-1}A_j$ also converges and to identify its limit--and for this I provided the hint: Cesàro. (Say, did you check the link?) –  Did Jun 18 at 12:43
    
Yes, I've checked the link. Thanks, Did, for details, now everything is clear. –  user44636 Jun 18 at 13:04

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