seek a direct method to show $(a_k)\in l^2$ 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. 
 A: 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 $n\ge k$
$(4)$ $\displaystyle \sum_{n=1}^\infty b_n\ge\sum_{n=k}^\infty b_n$
A: Hint: Cauchy condensation test.
As the sequence $\{a_k^2\}$ still is non-increasing, we have to check whether the series $\sum_n 2^na_{2^n}^2$ is convergent. As $\frac{a_k}{\sqrt k}$ is non-increasing, we have by condensation that the series $\sum_n2^n\frac{a_{2^n}}{2^{n/2}}$ is convergent.
A: For positive integers $n$, let $u(n) \in \ell^2$ such that $u(n)_j = 1$ for $j \le n$ and $0$ otherwise.  Then we can write $a = \sum_{j=1}^\infty b_j u(j)$ where $b_j = a_j - a_{j+1}$.  Now let $f(a) = \sum_{j=1}^\infty a_j/\sqrt{j}$.  We have 
$$f(u(n)) = \sum_{j=1}^n \frac{1}{\sqrt{j}} \ge \int_1^{n+1} \frac{dx}{\sqrt{x}} = 2 \sqrt{n+1} - 2 \ge c \sqrt{n} = c \|u(n)\|$$
for some constant $c > 0$ (in fact $c = 2 \sqrt{2}- 2$).  Thus 
$$f(a) = \sum_{j=1}^\infty b_j f(u(j)) \ge c^{-1} \sum_{j=1}^\infty b_j \|u(j)\| \ge c^{-1} \|a\|$$
Actually we should be a bit more careful with limits because $f$ is not a continuous linear functional (and is not defined for all $a$), but an appropriate limiting argument using partial sums should work.
