<s>[INST] If the partial sums $s_n $of $\sum a_n $ are bounded, show that the series $\sum \frac{a_n}{n} $ converges to $\sum \frac{s_n}{n (n+1)} $ My try:   $\left(\frac{1}{n}\right)_{n=1}^\infty $is decreasing and $\lim_{n\to\infty} \frac{1}{n}=0$ with $\sum a_n $is bounded so by Dirichlet's Test $\sum \frac{a_n}{n}$ is convergent.
To prove the given statement I am trying to show $|\sum \frac{a_n}{n}-\sum \frac{s_n}{n(n+1)}|$ tends to $0$.
\begin{align}
|\sum \frac{a_n}{n}-\sum \frac{s_n}{n(n+1)}|
&=\left|a_1 +\frac{a_2}{2} +\frac{a_3}{3} +\cdots-\frac{a_1}{1\cdot2} -\frac{a_1+a_2}{2\cdot3}-\frac{a_1+a_2+a_3}{3\cdot4} -\cdots\right|\\
&=|a_1-\frac{a_1}{1\cdot2} + \frac{a_2}{2} -\frac{a_1+a_2}{2\cdot3} +\frac{a_3}{3} -\frac{a_1+a_2+a_3}{3\cdot4} +\cdots|\\
&\leq\left|a_1-\frac{a_1}{1\cdot2}\right| + \left|\frac{a_2}{2} -\frac{a_1+a_2}{2\cdot3}\right| +\left|\frac{a_3}{3} -\frac{a_1+a_2+a_3}{3\cdot4}\right| + \cdots & (1)
\end{align}
Now From the $n$th term
$$\left|\frac{na_n-a_1-a_2-...-a_{n-1}}{n (n-1)}\right|\geq\left|\frac{na_n-M}{n(n-1)}\right|$$
($M$ is an upper bound of $\sum a_n$)
Also $\left|\frac{na_n+M}{n(n-1)}\right|$ is greater than the nth term. Taking $n$ tends to infinity we see that the $n$th term goes to $0$. Using this in $(1)$ we note that tail of both the series coincide and then $1$st series converges to $2$nd series. Am I correct?
 [/INST] Try the usual summation by parts as suggested in the comments.
Another approach:
$$\begin{align}\sum_{n=1}^N \frac{s_n}{n(n+1)} &= \sum_{n=1}^N \sum_{k=1}^n\frac{a_k}{n(n+1)} \\&= \sum_{n=1}^N \sum_{k=1}^N\frac{a_k\, 1_{k \leqslant n}}{n(n+1)} \\&= \sum_{k=1}^N \sum_{n=1}^N\frac{a_k\, 1_{k \leqslant n}}{n(n+1)} \\ &=  \sum_{k=1}^N a_k\sum_{n=k}^N\frac{1 }{n(n+1)}  \\ &= \sum_{k=1}^N a_k \left(\frac{1}{k} - \frac{1}{N+1} \right)\\ &= \sum_{k=1}^N \frac{a_k}{k} - \frac{s_N}{N+1} \end{align}$$
Now take limits as $N \to \infty$ using the fact that $s_N$ is bounded.
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