TITLE: Prove $\sum_1^{\infty} a_i^2$ convergent
QUESTION [6 upvotes]: Let$\{a_i\}$ be a decreasing sequence of positive number such than$\sum_{j=1}^{\infty}\frac{a_j}{\sqrt{j}}$ is convergent.
Prove $\sum_1^{\infty} a_i^2$ is convergent.
My attemption is using Abel test, clearly $\frac{a_j}{\sqrt{j}}$ is a decreasing sequence, and also $\sum_{j=1}^{\infty}\frac{a_j}{\sqrt{j}}$ is convergent, so $\{\frac{a_j}{\sqrt{j}}\}$ is bound, in order to use Abel test, we only need to prove $\sum \sqrt{j}a_i$ is convergent, but I do not know how to do it.

REPLY [3 votes]: Consider the set $S_x=\{j: \frac{a_j}{\sqrt{j}}\leq x^{j/2-1/2}\}$, $0<x<1$.
$$\sum_{S_x} a^2_j=\sum_{S_x} j\Big(\frac{a_j}{\sqrt{j}}\Big)^2\leq \sum_{S_x} jx^{j-1}<\sum_{j=1}^{\infty} \frac{d}{dx}(t^j)\Big{|}_{t=x}\\=\frac{d}{dx} \Big(\sum_{j=1}^{\infty}t^j\Big)\Big{|}_{t=x}<\infty,\space\space \text{for all $0<x<1$} $$