If $\sum{a_n}$ converges does that imply that $\sum{\frac{a_n}{n}}$ converges?

Question

I know if $\sum{a_n}$ converges absolutely then $\sum{\frac{a_n}{n}}$ converges since $0\le \frac{|a_n|}{n} \le |a_n| $ for all $n$ so it converges absolutely by the basic comparison test and therefore converges. However, I cannot prove the convergence of $\sum \frac{a_n}{n}$ if $\sum{a_n}$ converges but not absolutely even though I suspect it to be true. Can you give me a proof or a counterexample for this?

Answer 1

You might want to consider Dirichlet's test.

Answer 2

The problem can be handled by a summation by parts. Define $s_N:=\sum_{j=1}^N a_j$. Then for $M<N$, \begin{align} \sum_{j=M}^N\frac{a_j}j&=\sum_{j=M}^N\frac{s_j-s_{j-1}}j\\ &=\sum_{k=M}^N\frac{s_k}k-\sum_{k=M-1}^{N-1}\frac{s_k}{k+1}\\ &=\frac{S_N}N-\frac{s_{M-1}}M+\sum_{k=M}^{N-1}\frac{s_k}{k(k+1)}, \end{align} which yields the bound $$\left|\sum_{j=M}^N\frac{a_j}j\right|\leqslant \sup_k|s_k|\left(\frac 1M+\frac 1N+\sum_{k\geqslant M}\left(\frac 1k-\frac 1{k+1}\right)\right)\\ \\=\sup_k|s_k|\left(\frac 2M+\frac 1N\right).$$ This proves that the sequence $\left(\sum_{j=1}^Na_j/j\right)_{N\geqslant 1}$ is Cauchy, hence the convergence of the series $\sum_{j\geqslant 1}a_j/j$.

Answer 3

Obviously, this conclusion is correct. In mathematical analysis, the Abel discriminant of the series of real terms is described as follows.

If $a_n$ is monotonous and $\sum b_n$ converges, $\sum b_na_n$ converges.

Answer 4

Yes;

A theorem found in "Baby'' Rudin's book: If $\sum a_{n}$} converges and $\lbrace{ b_{n} \rbrace}$ monotonic and bounded then $\sum a_{n}b_{n}$ converges. See:

Prob. 8, Chap. 3 in Baby Rudin: If $\sum a_n$ converges and $\left\{b_n\right\}$ is monotonic and bounded, then $\sum a_n b_n$ converges.

Here, we take $b_{n} = \frac{1}{n}.$