convergence of $\sum\frac{a_{n}}{n}$ if $\sum_{k=1}^{n}a_{k}\le M*n^{r}$ where $r<1$ Show that if the partial sums $s_{n}$ of the series $\sum_{k=1}^{\infty}a_{k}$ satisfy $|s_{n}|\le M*n^{r}$ for some $r<1$, then the series $\sum_{n=1}^{\infty}\frac{a_{n}}{n}$ converges.
 A: OK, this time I think I got it. So using mentioned Abel's Lemma:
\begin{eqnarray}
 & \Big|\frac{a_{n+1}}{n+1}+\ldots+\frac{a_{m}}{m}\Big| \\
= & \Big|\frac{s_{m}}{m}-\frac{s_{n}}{n+1}+\sum_{k=n+1}^{m-1}\Big(\frac{1}{k}-\frac{1}{k+1}\Big)s_{k}\Big| \\
= & \Big|\frac{s_{m}}{m}-\frac{s_{n}}{n+1}+\sum_{k=n+1}^{m-1}\frac{s_{k}}{k(k+1)}\Big| \\
\le & \Big|\frac{s_{m}}{m}\Big|+\Big|\frac{s_{n}}{n+1}\Big|+\sum_{k=n+1}^{m-1}\Big|\frac{s_{k}}{k(k+1)}\Big| \\
\le & \frac{M}{m^{1-r}}+\frac{M}{(n+1)^{1-r}}+\sum_{k=n+1}^{m-
1}\frac{M}{k^{1-r}(k+1)} \\
\le & \frac{2M}{n^{1-r}}+M\sum_{k=n+1}^{m-1}\frac{1}{k^{2-r}} \\
\le & \frac{2M}{n^{1-r}}+M\sum_{k=n+1}^{\infty}\frac{1}{k^{2-r}} \\
\end{eqnarray}
Now, $r<1$ means that $2-r>1$. So, as p-series are convergent for $p>1$, then for given $\epsilon>0$, one can choose $N_{1}(\epsilon)$ such that $\sum_{k=n+1}^{\infty}\frac{1}{k^{2-r}}<\frac{\epsilon}{2M}$ whenever $n\ge N_{1}(\epsilon)$. Thus, if $N(\epsilon)=\max\Big(\big(\frac{4M}{\epsilon}\big)^{\frac{1}{1-r}},N_{1}(\epsilon)\Big)$, then the initial partial sum is smaller than $\epsilon$ for any $m>n\ge N(\epsilon)$, which means the series $\sum_{n=1}^{\infty}\frac{a_{n}}{n}$ converges by Cauchy criteria.
Any errors here?
A: Hint: Abel's transform, namely, write 
$$\sum_{n=1}^N\frac{a_n}n=\sum_{n=1}^N\frac{s_n}n-\sum_{n=1}^N\frac{s_{n-1}}n.$$
A: Define
$$
A_n=\sum_{k=1}^na_k\quad\Longrightarrow|A_n|\le Mn^r\tag{1}
$$
Then
$$
\begin{align}
\sum_{k=m}^n\frac{a_k}{k}
&=\sum_{k=m}^n\sum_{j=k}^\infty\frac{a_k}{j(j+1)}\\
&=\sum_{j=m}^\infty\sum_{k=m}^{\min(j,n)}\frac{a_k}{j(j+1)}\\
&=\sum_{j=m}^\infty\frac{A_{\min(j,n)}-A_{m-1}}{j(j+1)}\tag{2}
\end{align}
$$
Independent of $n$, we have
$$
\begin{align}
\sum_{j=m}^\infty\left|\frac{A_{\min(j,n)}}{j(j+1)}\right|
&\le\sum_{j=m}^\infty Mj^{r-2}\\
&\le\frac{M}{(1-r)m^{1-r}}\tag{3}
\end{align}
$$
And finally
$$
\begin{align}
\sum_{j=m}^\infty\left|\frac{A_{m-1}}{j(j+1)}\right|
&=\frac{|A_{m-1}|}{m}\\
&\le \frac{M}{m^{1-r}}\tag{4}
\end{align}
$$
Therefore, $(3)$ and $(4)$ show that $(2)\to0$ as $m,n\to\infty$. Thus, the series converges.
