Prove $\lim_{j\rightarrow \infty}\sum_{k=1}^{\infty}\frac{a_k}{j+k}=0$ I am only looking for a hint to start this exercise, not a full answer to the problem, please take this into consideration.
Suppose that $a_k \geq 0$ for $k$ large and that $\sum_{k=1}^\infty\frac{a_k}k$ converges. Prove that
$$\lim_{j\rightarrow \infty}\sum_{k=1}^{\infty}\frac{a_k}{j+k}=0$$
What I can see so far that may help is that, since $\sum_{k=1}^\infty a_k/k$ converges,
$\forall \epsilon > 0$, $\exists N\in \mathbb{N}$ such that $n\geq N\Rightarrow |\sum_{k=n}^\infty \frac{a_k}k|<\epsilon$, which is a result of the Cauchy Criterion. Once again, I am only looking for a hint to start this exercise, not a full proof.

I gave the hint in my comment.  For a full solution, read below:

Let $\epsilon>0$. 
Choose $N>1$ so that $a_j\ge 0$ for $j\ge N$ and such that $\sum\limits_{k=N}^\infty {a_k\over k}<\epsilon/2$.  Note that for $j>0$, we then have 
$$\tag{1}0\le \sum\limits_{k=N}^\infty {a_k\over k+j} \le\sum\limits_{k=N}^\infty {a_k\over k}<\epsilon/2.$$
So, we can make the tails $\sum\limits_{k=N}^\infty {a_k\over k+j}$ small (independent of $j$ in fact). 
Let's see how to make the remaining part of the sum, $\sum\limits_{k=1}^{N-1} {a_k\over j+k}$, small:
Let $M=\max\{|a_1|,\ldots |a_{N-1}| \}$. Choose $J> M(N-1)/(2\epsilon)$.
Then for $j\ge J$:
$$\tag{2}\Bigl| \,\sum\limits_{k=1}^{N-1} {a_k\over j+k}\,\Bigr|\le 
\sum\limits_{k=1}^{N-1} {M\over J }=(N-1)M\cdot{1\over J}<\epsilon/2.
$$
Using $(1)$ and $(2)$, we have for $j>J $:
$$
\Bigl|\,\sum_{k=1}^\infty {a_k\over j+k}\,\Bigr|
\le
\Bigl|\,\sum_{k=1}^{N-1} {a_k\over j+k}\,\Bigr|+
\Bigl|\,\sum_{k=N }^\infty {a_k\over j+k}\,\Bigr|
\le{\epsilon\over2}+
{\epsilon\over2}=\epsilon.
$$
Since $\epsilon$ was arbitrary, the result follows.
