seek a direct method to show $(a_k)\in l^2$

Solution 1:

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 $n\ge k$

$(4)$ $\displaystyle \sum_{n=1}^\infty b_n\ge\sum_{n=k}^\infty b_n$

Solution 2:

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.