TITLE: Does $\big|\sum_{k=1}^na_k\big|\leq\sqrt{n}$ imply the convergence of $\sum_{1}^\infty \frac{a_k}{k}$?
QUESTION [5 upvotes]: Suppose $\big|\sum_{k=1}^na_k\,\big|\leq\sqrt{n}$ for all $n\geq 1$. Show that $\sum_{1}^\infty \frac{a_k}{k}$ converges. 

 Summation by part doesn't help. ([Added: Thanks to the comments, I should have made my thought clear so that it would not cause confusion. It is not that "Summation by part doesn't help", but that "I don't see how 'summation by part' might work".]) I tried several convergence tests without any progress. The assumption gives an estimate for the Cesaro mean:
$$
\left|\frac{\sum_{k=1}^n a_k}{n}\right|\leq \frac{1}{\sqrt n}.
$$
It seems that if I can get
$$
a_n\sim \left|\frac{\sum_{k=1}^n a_k}{n}\right|
$$
then things will be done since 
$$
\sum\frac{1}{\sqrt k \cdot k}
$$
converges. How shall I go on?

REPLY [7 votes]: Thanks to all the comments above, I just found that I had been close to the answer. Summation by part would definitely work. 
Let $A_n=\sum_{k=1}^n a_k$ and $b_k=\frac 1k$. Then 
$$
\begin{align}
\sum_{k=1}^na_kb_k&=\sum_{k=1}^{n-1}A_kb_k-\sum_{k=1}^{n-1}A_kb_{k+1}+A_nb_n\\
&=\sum_{k=1}^{n-1}A_k\left(\frac{1}{k}-\frac{1}{k+1}\right)+A_n\frac{1}{n}\\
&=\sum_{k=1}^{n-1}A_k\frac{1}{k(k+1)}+A_n\frac{1}{n}
\end{align}
$$
The first term is convergent since
$$
\frac{A_k}{k(k+1)}\sim\frac{1}{k^{3/2}}.
$$
So is the second term:
$$
A_n/n\to 0.
$$