Proving that the given sequence converges I have been having trouble proving the following result, which I have written below.
Let $(a_{n=1}^\infty)$ be a sequence of positive real numbers and let $A_n$ = $\sum\limits_{i=1}^n a_i$ Prove that if  $\sum\limits_{n=1}^\infty \dfrac{a_n}{A_n}$ < $\infty$, then $\sum\limits_{n=1}^n a_n$ < $\infty$.
I have been given the following hint. 
Assume that$\sum\limits_{n=1}^\infty a_n$ = $\infty$. Let 0 < $\epsilon$ < 1/2. Then for a sufficiently large N $\geq$ 2 and all n > N, $\epsilon$ >$\sum\limits_{j=N}^\infty \dfrac{a_j}{A_j}$  $\geq$ $\sum\limits_{j=N}^n \dfrac{a_j}{A_j}$.
I am unsure how to proceed. Thanks for any help in advance.
 A: You get 
$$\varepsilon > \sum_{j=N}^n \frac{a_j}{A_j}\ge \frac{1}{A_n}\sum_{j=N}^n a_j$$
for every $n>N$, since $A_j$ is monotonically increasing. 
Can you finish the proof with this hint?
A: By the hint, for all $n\ge N$ we have $\frac 1 {A_n}\sum_{j=N}^na_j < \varepsilon$, hence $\sum_{j=N}^na_j < \varepsilon\sum_{j=1}^na_j$. This is equivalent to $(1-\varepsilon)\sum_{j=N}^na_j < \varepsilon\sum_{j=1}^{N-1}a_j$. Therefore, $\sum_{j=N}^na_j < \frac\varepsilon{1-\varepsilon}\sum_{j=1}^{N-1}a_j$ for all $n\ge N$. Letting $n\to\infty$ proves the claim.
A: The $\epsilon$ in the hint seems confusing.  All we really need is that since $\sum\limits_{n=1}^\infty\frac{a_n}{A_n}$ converges, there is an $N$ so that $\sum\limits_{n=N}^\infty\frac{a_n}{A_n}\le\frac12$. Therefore, since $A_n$ is increasing, for any $m\ge N$,
$$
\begin{align}
A_m-A_{N-1}
&=\sum_{n=N}^ma_n\\
&\le\sum_{n=N}^m\frac{A_m}{A_n}a_n\\
&\le\sum_{n=N}^\infty\frac{a_n}{A_n}A_m\\
&\le\frac12A_m\tag{1}
\end{align}
$$
$(1)$ is equivalent to
$$
A_m\le2A_{N-1}\tag{2}
$$
Taking the limit of $(2)$ as $m\to\infty$ gives
$$
\sum_{n=1}^\infty a_n\le2A_{N-1}\tag{3}
$$
