# Show that if the partial sums $s_n$ of $\sum_{k=1}^\infty a_k$ satisfy $|s_n|\leq Mn^r$ for some $r<1$, then $\sum_{n=1}^\infty a_n/n$ converges.

The question is:

Show that if the partial sums $s_n$ of the series $\sum\limits_{k=1}^\infty a_k$ satisfy $\vert{s_n}\vert\leq Mn^r$ for some $r<1$, then the series $\sum\limits_{n=1}^\infty a_n/n$ converges.

My Attempt:

Let $m>n, m,n\in \mathbb{N}$ By Abel's Lemma, $$\sum\limits_{k=n+1}^m \frac{a_k}k=\frac{s_m}m-\frac{s_n}{n+1}+\sum\limits_{k=n+1}^{m-1} \left(\frac1k-\frac1{k+1}\right)s_k$$

Now I have to use the condition $\vert{s_n}\vert\leq Mn^r$ but don't know how to do so.. Could somebody help me with the solution? Or if there is a better way of doing it, could you teach me it?

• Presumably you mean $a_k/k$ on the left side of the last equation. – Thomas Andrews Jun 16 '14 at 19:25
• And $\frac{1}{k} - \frac{1}{k+1}$ in the sum on the right. – Daniel Fischer Jun 16 '14 at 19:27

Hint: $$\left|\frac{1}{k}-\frac{1}{k+1}\right|<\frac{1}{k^2}$$
$\left|s_{n}\right|=\left|\sum\limits_{k=1}^{n}a_{k}\right|\leq M\cdot n^{r}\quad\Rightarrow\quad \left|\frac{1}{n}\sum\limits_{k=1}^{n}a_{k}\right|\leq M\cdot n^{r-1}$, where $r-1<0$.
Hence $\left|\frac{1}{n}\sum\limits_{k=1}^{n}a_{k}\right|\to 0$, if $n\to\infty$. Now use the Cauchy criterion, or something like that.
• it is $\sum_{n=1}^\infty \frac 1n a_n$ not $\lim_{n\to\infty} \frac1n \sum_{i=1}^n a_n$! – daw Jun 16 '14 at 19:46