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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.

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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. –  David Mitra Feb 14 '13 at 2:08
    
Doesn't that inequality assume $j>0$? –  kaiserphellos Feb 14 '13 at 2:16
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Yes, but this is safe to assume (we are taking the limit as $j$ approaches $\infty$) –  David Mitra Feb 14 '13 at 2:24
    
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 $j>0$. –  kaiserphellos Feb 14 '13 at 2:25

4 Answers 4

up vote 2 down vote accepted

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.

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

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Hint: Lebesgue dominated convergence theorem.

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

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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! –  kaiserphellos Feb 14 '13 at 2:18
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@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 $\mathbb{N}$. –  1015 Feb 14 '13 at 2:23

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.

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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 $0$. –  kaiserphellos Feb 14 '13 at 2:44
    
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... –  DonAntonio Feb 14 '13 at 2:51
    
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 $\sum_{k=1}^j {a_k\over j+k}$? –  David Mitra Feb 14 '13 at 2:56
    
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. –  DonAntonio Feb 14 '13 at 2:58
    
How is it an equality? You would need to change the index of $a_k$ for that. –  David Mitra Feb 14 '13 at 3:01

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