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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?

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    $\begingroup$ Hint: write $a_n = s_{n} - s_{n-1}$ in $q_N = \sum_{n=1}^N \frac{a_n}{n}$. $\endgroup$ – Gribouillis Sep 9 '17 at 17:25
  • $\begingroup$ @RRL Then both of the series coincides for large n,Isn't it ? $\endgroup$ – Subhajit Saha Sep 9 '17 at 18:21
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    $\begingroup$ You've shown both series converge, so it is always the case that the difference of terms tends to $0$ since each term tends to $0$. No estimate is necessary. That alone is not enough to show they converge to the same value. $\endgroup$ – RRL Sep 9 '17 at 18:40
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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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