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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.

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    $\begingroup$ I think you can't follow this way because if you take $a_j = 1/j$ You have a decreasing sequence of positive numbers such that $ \sum(a_j)/\sqrt{j}$ converges but $\sum \sqrt{j}a_j$ does not $\endgroup$ – jJjjJ Dec 9 '16 at 8:05
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You just need to apply to $b_j=\frac{a_j}{\sqrt{j}}$ the result that if $b_j$ is positive and decreasing and if $\sum b_j$ converges, then $\lim\limits_{j \to \infty} jb_j = 0$. (This has been proved many times on this site. See for example here.)

Since $\frac{a_j}{\sqrt{j}}$ is decreasing (because $a_j$ is positive and decreasing) and $\sum \frac{a_j}{\sqrt{j}}$ converges we must have $\sqrt{j} a_j = j \frac{a_j}{\sqrt{j}} \to 0$ as $j \to \infty.$

Hence, for sufficiently large $j$, we have $\sqrt{j} a_j < 1$ and

$$a_j^2 = \sqrt{j}a_j \frac{a_j}{\sqrt{j}} \leqslant \frac{a_j}{\sqrt{j}}. $$

Thus, $\sum a_j^2$ must converge by the comparison test.

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Consider the set $S_x=\{j: \frac{a_j}{\sqrt{j}}\leq x^{j/2-1/2}\}$, $0<x<1$.

$$\sum_{S_x} a^2_j=\sum_{S_x} j\Big(\frac{a_j}{\sqrt{j}}\Big)^2\leq \sum_{S_x} jx^{j-1}<\sum_{j=1}^{\infty} \frac{d}{dx}(t^j)\Big{|}_{t=x}\\=\frac{d}{dx} \Big(\sum_{j=1}^{\infty}t^j\Big)\Big{|}_{t=x}<\infty,\space\space \text{for all $0<x<1$} $$

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