2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ For $n$ fixed, can you make $\sum\limits_{k=1}^{n-1}{a_k\over j+k}$ small by taking $j$ large? Also, note $\sum\limits_{k=n}^{\infty}{a_k\over j+k}\le\sum\limits_{k=n}^{\infty}{a_k\over k}$ for $n$ sufficiently large. $\endgroup$ – David Mitra Feb 14 '13 at 2:08
  • $\begingroup$ Doesn't that inequality assume $j>0$? $\endgroup$ – kaiserphellos Feb 14 '13 at 2:16
  • 1
    $\begingroup$ Yes, but this is safe to assume (we are taking the limit as $j$ approaches $\infty$) $\endgroup$ – David Mitra Feb 14 '13 at 2:24
  • $\begingroup$ Thank you, I believe that this insight is what I needed to complete the proof. I suppose I was really just unsure whether or not we could assume $j>0$. $\endgroup$ – kaiserphellos Feb 14 '13 at 2:25
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

Hint: Lebesgue dominated convergence theorem.

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

$\endgroup$
  • $\begingroup$ Yes, this theorem has not been stated in the course yet, therefore I cannot use it to prove the proposition. I'll definitely look into this theorem though! $\endgroup$ – kaiserphellos Feb 14 '13 at 2:18
  • 1
    $\begingroup$ @ChristopherWashington I assumed you had not seen it yet. But I couldn't resist. This theorem is so powerful. You'll see it stated for functions on a measure space. But series are such functions with the appropriate measure on $\mathbb{N}$. $\endgroup$ – Julien Feb 14 '13 at 2:23
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I believe that this would work, however, wouldn't we need to use a sufficiently large $n$ for the starting index? Since in the problem statement we only assume $a_k\geq 0$ for large k, so $a_1, a_2,...,a_{n-1}$ could be less than $0$. $\endgroup$ – kaiserphellos Feb 14 '13 at 2:44
  • $\begingroup$ Indeed, but as noted in the answer we can "throw" away the first $\,n\,$ elements of the sequence $\,\{a_n\}\,$ without changing the fact that $\,\sum\frac{a_k}{k}\,$ converges... $\endgroup$ – DonAntonio Feb 14 '13 at 2:51
  • $\begingroup$ I don't see how your inequality holds. If the second sum was from $k=1$ to $\infty$, it would be fine. What happened to $\sum_{k=1}^j {a_k\over j+k}$? $\endgroup$ – David Mitra Feb 14 '13 at 2:56
  • $\begingroup$ Well, the inequality in fact is a equality...:) , and why would we care about your sum? I don't think it fits what the OP asked. $\endgroup$ – DonAntonio Feb 14 '13 at 2:58
  • $\begingroup$ How is it an equality? You would need to change the index of $a_k$ for that. $\endgroup$ – David Mitra Feb 14 '13 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.