Suppose that $\sum_{j=1}^na_j0$. Show that $\sum_{n=1}^\infty \frac{a_n}{n}$ converges. 
Suppose for some $\epsilon>0$ that the partial sums of a series are $\sum_{j=1}^na_j<n^{1-\epsilon}$. I want to show that $\sum_{n=1}^\infty\frac{a_n}{n}$ converges.

I am not sure I know a test to apply to figure out the convergence of this series.
Abel's test does not work because $n^{1-\epsilon}$ is not a uniform bound. It's not obvious how I would use Kummer's test because I don't know anything about the ratio of successive terms.
The only thing I can think is that surely $\displaystyle a_n<\sum_{j=0}^na_j<n^{1-\epsilon}$, so then
 $\displaystyle\sum_{n=1}^\infty\frac{a_n}{n}<\sum_{n=1}^\infty\frac{n^{1-\epsilon}}{n}$ but this isn't very helpful since the one on the right diverges :(
The only other thing I tried was using summation by parts. Let $b_n=\frac{1}{n}$ then 
$$\sum_{j=1}^Na_jb_j=b_N\sum_{j=1}^Na_j + \sum_{n=1}^{N-1}\left(\sum_{j=1}^na_j(b_n-b_{n+1})\right)$$
$$\sum_{j=1}^N\frac{a_j}{j}<\frac{N^{1-\epsilon}}{N}+\sum_{n=1}^{N-1}n^{1-\epsilon}\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ <\frac{1}{N^\epsilon}+\sum_{n=1}^{N-1}\left(1-\frac{n}{n+1}\right)$$
But then again, the series on the right is divergent when we let $N\rightarrow \infty$ so this was not helpful.
Is there a clear test to use? Have I made a critical error?
 A: Suppose
$$
\sum_{k=1}^na_k\le n^{1-\epsilon}\tag1
$$
where $a_n\ge0$. Then
$$
\begin{align}
\sum_{n=1}^\infty\frac{a_n}n
&=\sum_{n=1}^\infty\sum_{k=n}^\infty\left(\frac1k-\frac1{k+1}\right)a_n\tag2\\
&=\sum_{k=1}^\infty\sum_{n=1}^k\frac1{k(k+1)}\,a_n\tag3\\
&\le\sum_{k=1}^\infty\frac1{k^2}\,k^{1-\epsilon}\tag4\\
&=\sum_{k=1}^\infty k^{-1-\epsilon}\tag5
\end{align}
$$
Explanation:
$(2)$: write $\frac1n$ as the sum of a telescoping series
$(3)$: change the order of summation and $\frac1k-\frac1{k+1}=\frac1{k(k+1)}$
$(4)$: apply $(1)$ and $\frac1{k(k+1)}\lt\frac1{k^2}$
$(5)$: simplify
The sum in $(5)$ converges for $\epsilon\gt0$.
A: Denote $s_n=\sum_{k=1}^n a_j$
By assumption we have ($\epsilon>0$):
$$
s_n\leq n^{1-\epsilon} \quad (1)
$$
Using summation by parts with $a_n=s_n-s_{n-1}$ and $\tfrac{1}{n}-\tfrac{1}{n+1}=\tfrac{n}{n+1}$ we get (the upper boundary term can be neglected since $s_n/n$ goes to zero for large $n$ by assumption (1)). 
$$
\sum_{n=1}^{\infty}\frac{a_n}{n}=\sum_{n=2}^{\infty}\frac{s_n}{n(n+1)}+a_1
$$
or by (1)
$$
\sum_{n=1}^{\infty}\frac{a_n}{n}\leq \sum_{n=2}^{\infty}\frac{1}{n^{\epsilon}(n+1)}+a_1<\infty
$$
which means that the series converges
