# Limit of sequence

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.

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Got something from my answer? –  Did Dec 20 '14 at 8:29

Hint: Apply Cesàro to $$b_n=A_n-\frac1n\sum_{j=1}^{n-1}A_j,\qquad A_j=\sum_{k=1}^j\frac{a_k}k.$$
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 '14 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 '14 at 12:43