Show the series $\sum\limits^{\infty}_{n=1}{\frac{a_n}{n}}$ converges 
Suppose the sequence $\{a_n\}_{n=1}^{\infty}$ satisfies
  $$\mid\sum\limits^{n}_{k=1}{a_{k}}\mid\leq C\sqrt{n} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space n=1, 2, 3, \cdots$$
  where $C$ is a fixed (but arbitrary) number. Prove that the series
  $$\sum\limits^{\infty}_{n=1}{\frac{a_n}{n}}$$
  converges.

My attempt: Suppose $b_n:= \frac{1}{n}$; Abel's lemma on summation by parts gives
$$\sum\limits^{k}_{n=1}{\frac{a_n}{n}}=\sum\limits^{k-1}_{n=1}{[\sum\limits_{i=1}^{n}{a_i}\cdot(b_n-b_{n+1})] + \sum\limits_{i=1}^{k}{a_i}\cdot b_k}$$
$$<\sum\limits^{k-1}_{n=1}{\mid\sum\limits_{i=1}^{n}{a_i}\mid\cdot(b_n-b_{n+1}) +\mid \sum\limits_{i=1}^{k}{a_i}\mid \cdot b_k}$$
$$\le\sum\limits_{n=1}^{k}[C\sqrt{n}\cdot{(\frac{1}{n}-\frac{1}{n+1}})]+C\sqrt{k}\cdot \frac{1}{k+1}$$
$$=\sum\limits_{n=1}^{k}{\frac{C\sqrt{n}}{n(n+1)}}+\frac{C\sqrt{k}}{k+1}.$$
Since $k\rightarrow\infty$, therefore
$$\frac{C\sqrt{k}}{k+1}\rightarrow 0.$$
Moreover, for the sigma notation, since
$$\frac{C\sqrt{n}}{n(n+1)}<\frac{C\sqrt{n}}{n^2}=\frac{C}{n^\frac{3}{2}}$$
Above is a $p$-series with $p=\frac{3}{2}>1$, hence the series $\sum\limits_{n=1}^{k}{\frac{C\sqrt{n}}{n(n+1)}}$ converges. Even though $\sum\limits_{n=1}^{k}{\frac{C\sqrt{n}}{n(n+1)}}$ converges and $\frac{C\sqrt{k}}{k+1}$ approaches to $0$ for $k\rightarrow\infty$, but they do not imply the series $\sum\limits^{\infty}_{n=1}{\frac{a_n}{n}}$ converges.
My proof seems not yet completed. How do I continue it?
 A: Let
$$
s_n=\sum_{k=1}^na_n
$$
then $|s_n|\le C\sqrt{n}$ and
$$
\begin{align}
\sum_{k=1}^n\frac{a_k}k
&=\sum_{k=1}^n\frac{s_k-s_{k-1}}k\\
&=\sum_{k=1}^n\frac{s_k}k-\sum_{k=0}^{n-1}\frac{s_k}{k+1}\\
&=\sum_{k=1}^ns_k\left(\frac1k-\frac1{k+1}\right)+\frac{s_n}{n+1}\\
&=\sum_{k=1}^n\underbrace{\frac{s_k}{k(k+1)}}_{\le C\frac{\sqrt{k}}{k^2}}+\underbrace{\ \ \frac{s_n}{n+1}\ \ }_{\le C\frac{\sqrt{n}}{n\vphantom{k^2}}}
\end{align}
$$
The sum converges and the extra term vanishes.

Comment brought into the answer
Since it does clarify the answer, I will bring the following comment into the answer.

The sequence of partial sums $\sum\limits_{k=1}^n\frac{a_k}k$ is the same as the sequence $\sum\limits_{k=1}^n\frac{s_k}{k(k+1)}+\frac{s_n}{n+1}$ the latter sum converges absolutely, hence the partial sums are Cauchy and the extra term converges to $0$ hence it is Cauchy. The sum of two Cauchy sequences is Cauchy. Since the original sequence of partial sums is Cauchy, it converges.

A: In order to prove the convergence, it is enough to show that this sequence is a Cauchy-sequence (another way to prove the convergence is given by robjohn), because $\mathbb{R}$, resp. $\mathbb{C}$ is complete. Write $A_k := \sum_{i=1}^k a_i$ and $B_k := \sum_{i=1}^k a_i/i$. Note that $a_k = A_{k} - A_{k-1}$ Then for $n>m$
\begin{align}
 |B_n - B_m| = \left| \sum_{i=m+1}^n \frac{a_i}{i} \right| &= \left| \sum_{i=m+1}^n \frac{A_i -A_{i-1}}{i} \right| \\
& =  \frac{|A_m|}{m+1} + \frac{|A_n|}{n} +  \left| \sum_{i=m+1}^{n-1} A_i \left(\frac{1}{i} - \frac{1}{i-1} \right) \right|
\end{align}
Since $(i^{-1}-(i-1)^{-1})= -(i(i-1))^{-1}$ Now we can use the bound to get the bound
$$\frac{2C}{\sqrt{m}}+ C \sum_{i=m+1}^{n-1} \frac{1}{(i-1) \sqrt{i}}.$$
The last sum can be estimated by $\int_{m-1}^\infty \frac{1}{x^{3/2}} d x < \frac{3}{2} (m-1)^{-1/2}.$ All in all, we see that $$|B_n-B_m| \le \frac{3C}{(m-1)^{1/2}},$$
i.e. this is a Cauchy-sequence as claimed.
Note that this argument works also if we only require that $$|\sum_{k=1}^n a_k| \le C n^{1-\varepsilon}.$$
Moreover, we cannot expect that $\sum_{k=1}^\infty a_k/k$ converges absolutely. Just take $a_k = (-1)^k$, then $$|\sum_{k=1}^n a_k| \le 1 \le \sqrt{n},$$ but $$\sum_{k=1}^\infty \frac{|a_k|}{k} = \sum_{k=1}^\infty \frac{1}{k} =\infty.$$
