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If $\sum\limits_{n=1}^\infty a_n$ converges, is the series $\sum\limits_{n=1}^\infty\dfrac{a_n}{n}$ also convergent?

I've found an answer online that said the second series is convergent,but it did't give the proof,I have no idea how to prove it is true or not.


marked as duplicate by StubbornAtom, Ali Caglayan, the_candyman, Adrian Keister, Martin Argerami Jan 28 at 14:56

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    $\begingroup$ Check Abel's test: en.wikipedia.org/wiki/Abel%27s_test $\endgroup$ – DonAntonio Dec 13 '17 at 8:29
  • $\begingroup$ if the $a_n's$ are posistive then use the coparaison test $\endgroup$ – Guy Fsone Dec 13 '17 at 8:35
  • $\begingroup$ Abel's test definitely can solve this problem,thank you guys. $\endgroup$ – burg1ar Dec 13 '17 at 9:00

Set $$ s_n=a_1+\cdots+a_n, \quad n\in\mathbb N. $$ Then $\{s_n\}$ converges, say to $s\in\mathbb R$. Next $$ \sum_{k=1}^n\frac{a_k}{k}=\sum_{k=1}^n\frac{s_k-s_{k-1}}{k}= \sum_{k=1}^n\frac{s_k}{k}- \sum_{k=2}^n\frac{s_{k-1}}{k}= \sum_{k=1}^n\frac{s_k}{k}- \sum_{k=1}^{n-1}\frac{s_{k}}{k+1}\\= \sum_{k=2}^ns_k\left(\frac{1}{k}-\frac{1}{k+1}\right)+\frac{s_n}{n} = \sum_{k=2}^n\frac{s_k}{k(k+1)}+\frac{s_n}{n}. $$ Clearly, $\dfrac{s_n}{n}\to 0$, and $$ \sum_{k=2}^n\frac{|s_k|}{k(k+1)}\le M\sum_{k=2}^n\frac{1}{k(k+1)}< \frac{M}{2}, $$ where $M$ is an upper bound of $\{|s_n|\}$, and hence $$ \sum_{k=2}^n\frac{s_k}{k(k+1)} $$ converges, due to the Comparison Test.

  • $\begingroup$ Could we not have just used the direct comparison test with $0\leq \frac{a_k}{k} \leq a_k \Rightarrow \sum^n \frac{a_k}{k} \leq \sum^n a_k $? I feel like I'm missing a trick here. $\endgroup$ – Jam Dec 13 '17 at 9:04
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    $\begingroup$ @Jam What you say would work if we knew that $\{a_n\}$ had only non-negative terms. All this trouble in my answer is to cover the case of general $\{a_n\}$, even the case of conditionally convergent series. $\endgroup$ – Yiorgos S. Smyrlis Dec 13 '17 at 9:09
  • $\begingroup$ Of course; thanks. $\endgroup$ – Jam Dec 13 '17 at 9:13

Dirichlet's test states that if the partial sums of $\{a_n\}_{n\geq 1}$ are bounded and $\{b_n\}_{n\geq 1}$ is decreasing towards zero then $\sum_{n\geq 1}a_n b_n$ is convergent (by summation by parts). If $\sum_{n\geq 1}a_n$ is convergent its partial sums are clearly bounded, and $\left\{\frac{1}{n}\right\}_{n\geq 1}$ is blatantly decreasing towards zero. It follows that $$ \sum_{n\geq 1}a_n\text{ is convergent}\quad\Longrightarrow\quad \sum_{n\geq 1}\frac{a_n}{n}\text{ is convergent}.$$ Kronecker's lemma provides a sort of converse statement: $$ \sum_{n\geq 1}\frac{a_n}{n}\text{ is convergent}\quad\Longrightarrow\quad a_1+a_2+\ldots+a_n = o(n).$$


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