Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 (

share|cite|improve this question
Got something from my answer? – Did Dec 20 '14 at 8:29

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

share|cite|improve this answer
Can anyone give more detail, please? – user44636 Jun 18 '14 at 10:38
Sure--as soon as you explain what is blocking you. – Did Jun 18 '14 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 '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
Yes, I've checked the link. Thanks, Did, for details, now everything is clear. – user44636 Jun 18 '14 at 13:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.