# If $\sum\limits_{n=1}^\infty a_n$ is convergent, then so is $\sum\limits_{n=1}^\infty\frac{a_n}{n}$ [duplicate]

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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 ArgeramiJan 28 at 14:56

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

## 2 Answers

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.

• 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. – Jam Dec 13 '17 at 9:04
• @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. – Yiorgos S. Smyrlis Dec 13 '17 at 9:09
• Of course; thanks. – 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: $$\sum_{n\geq 1}\frac{a_n}{n}\text{ is convergent}\quad\Longrightarrow\quad a_1+a_2+\ldots+a_n = o(n).$$