Is $\sum_{n=1}^\infty {a_n \over \sqrt{n}}$ (absolutely) convergent?

Question

Let $\sum_{n=1}^\infty a_n$ be a convergent series. Does this mean that:

1) $\sum_{n=1}^\infty {a_n \over \sqrt{n}}$ is convergent?

2) $\sum_{n=1}^\infty {a_n \over \sqrt{n}}$ is absolutely convergent?

If $\sum_{n=1}^\infty a_n$ is convergent, then we know it has a constrained sequence of partial sums. At the same time ${1 \over \sqrt{n}} \to 0$ monotonically which allows us to use Dirichlet's test to conclude that $\sum_{n=1}^\infty {a_n \over \sqrt{n}}$ is convergent, so my answer to 1) would be "yes." My sense about 2) is that this doesn't have to be the case but I'm struggling to find a counterexample? Any comments?

Answer 1

Denote $S_n:=\sum_{i=1}^na_i$ and $S_0:=0$. Then for any integers $N\leqslant M$, \begin{align} \sum_{n=N}^M\frac{a_n}{\sqrt n}&= \sum_{n=N}^M\frac{S_n-S_{n-1} }{\sqrt n}\\ &= \sum_{n=N}^M\frac{S_n }{\sqrt n}-\sum_{l=N-1}^{M-1} \frac{S_{l} }{\sqrt{l+1}} \\ &=\frac{S_M}{\sqrt M}-\frac{S_{N-1}}{\sqrt N} + \sum_{l=N}^{M-1} S_{l} \left(\frac 1{\sqrt l}- \frac{1 }{\sqrt{l+1}} \right). \end{align}
Using boundedness of the sequence $\left(S_l\right)_{l\geqslant 1}$, we derive that (1) holds.

Sergei Golovan gave a counter-example for (2): take $a_n:=(-1)^n/\sqrt n$. In this case, the series $\sum_{n=1}^\infty a_n$ converges but $\left|a_n /\sqrt n\right|=1/n$.