TITLE: seek a direct method to show $(a_k)\in l^2$
QUESTION [10 upvotes]: Let $\{a_i\}, i=1, 2, \cdots$ be a nonincreasing sequence of positive numbers, and suppose $\sum_{k=1}^{\infty}\frac{a_k}{\sqrt{k}}<\infty~$, show that $\sum_{k=1}^{\infty}a_k^2<\infty.$
Could you give a direct proof by finding an inequality of the type $\sum_{k=1}^{\infty}a_k^2\leq f<\infty$, where $f$ is some expression related to the already known number $\sum_{k=1}^{\infty}\frac{a_k}{\sqrt{k}}$?
Thanks in advance!

Remark:
I have solved this problem, but my method is indirect, I have used the uniformly bounded principal and the inequality $lim_{k\to\infty}\frac{1}{\sqrt{k}}\sum_{j=1}^ka_j=0,\forall (a_k)\in l^2$, so I really want to know how to solve the problem by finding a direct inequality.

REPLY [7 votes]: Set $b_n=\frac{a_n}{\sqrt n}$, then the result desired is equivalent to
$$
\begin{align}
\sum_{n=1}^\infty nb_n^2
&=\sum_{n=1}^\infty\sum_{k=1}^n b_n^2\tag{1}\\
&=\sum_{k=1}^\infty\sum_{n=k}^\infty b_n^2\tag{2}\\
&\le\sum_{k=1}^\infty b_k\sum_{n=k}^\infty b_n\tag{3}\\
&\le\left(\sum_{n=1}^\infty b_n\right)^2\tag{4}
\end{align}
$$
where $b_n$ is a monotonically decreasing positive sequence.
Explanation of Steps:
$(1)$ $\displaystyle n=\sum_{k=1}^n1$
$(2)$ change order of summation
$(3)$ $b_k\ge b_n$ for $n\ge k$
$(4)$ $\displaystyle \sum_{n=1}^\infty b_n\ge\sum_{n=k}^\infty b_n$