Prove $\lim_{j\rightarrow \infty}\sum_{k=1}^{\infty}\frac{a_k}{j+k}=0$

Question

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.

Answer 1

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.

Answer 2

Hint: prove the series converges uniformly using M-test.

Answer 3

Hint: Lebesgue dominated convergence theorem.

For an elementary proof not using this big hammer, see David Mitra's comment!

Answer 4

I think in this case you can use as basic a thing as the squeeze theorem.

First, we can assume $\,a_k\ge 0\,\,\,\forall\,k\in\Bbb N\,$ and nothing about convergence changes (though, of course, the sum of the infinite convergent series may change), so

$$0\le \sum_{k=1}^\infty\frac{a_k}{j+k}\le\sum_{k=j+1}^\infty\frac{a_k}{k}\xrightarrow [j\to\infty]{}0$$

since we know the tail of a convergent series converges to zero.