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I had this problem on my last exam in calculus, I still don't know, how to solve this, so I would appreciate your help.

Let $a_{ij} \in \mathbb{R}$ for $i,j \in \mathbb{N}$. Suppose, that for every $j$ series $S_j = \sum_{i=1}^{\infty}{a_{ij}}$ is absolutely convergent and, for every $i$ exists finite limit (limit $\not= \pm\infty$) $c_i=\lim_{j\to\infty}{a_{ij}}$.

(a) Prove, that if exists such absolutely convergent series $\sum_{i=1}^{\infty}{b_i}$ that $|a_{ij}| \leq |b_i|$ for every $i,j$ then $S_j \to S = \sum_{i=1}^{\infty}{c_i}$, $j \to \infty$

(b) Is argument in (a) true without assumption that series $\sum{}b_i$ with properties given in (a) exists?

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My first reaction is that this is the discrete case of the dominated convergence theorem (i.e., for the counting measure). This sounds like a bit much for a "calculus" exam though, and a more elementary explanation would be desirable. Probably someone else will give one shortly... –  Pete L. Clark Dec 19 '10 at 13:40
    
@Pete: the problem I see with going through DCT is that one possible textbook presentation is to prove DCT from Fatou's lemma, and prove Fatou's lemma from the monotone convergence theorem, and the MCT is proven from approximation by step functions, and the step-function approximates are shown to converge using (a variant of) the result of this exercise. :) –  Willie Wong Dec 19 '10 at 14:56
    
@Willie Wong: I see other problems besides that, which is why it was a remark rather than an answer. Thanks for your answer though, which was very nice, even if you "wasted" $\frac{\epsilon}{12}$. :) –  Pete L. Clark Dec 19 '10 at 21:25

1 Answer 1

up vote 5 down vote accepted

First let us look at part (b) to extract some ideas about proving part (a).

Consider the double sequence $a_{ij} = \delta_{ij}$, that is, $a_{ij} = 1$ if $i = j$ and $a_{ij} = 0$ otherwise. Then $S_j = 1$ for any $j$. On the other hand, $c_i = 0$ for any $i$. So $S = \sum c_i = 0 \neq \lim S_j$. So without the dominating sequence $b_i$ the statement is false.

(Like Pete says, this is another reflection of something in the continuous case, namely the principle of weak convergence and also Fatou's lemma.)

Now we prove part (a). Fix $\epsilon > 0$. Let $N$ be chosen large enough such that $\sum_{i = N}^\infty |b_i| < \epsilon/4 $, this $N$ exists since $(b_i)$ is absolutely convergent. Consider the partial sums $S_{j,N} = \sum_{i = 1}^{N-1} a_{ij}$. We can then pick $M$ sufficiently large such that for all $j > M$ and $i < N$, $|a_{ij} - c_i| < \epsilon / (4N)$.

So in particular, $|S_{j,N} - S_N| < \epsilon / 4$ for $j > M$ where $S_N = \sum_{i = 1}^{N-1} c_i$.

Now, by assumption of convergence, and the domination of $(b_i)$, you have $|c_i| < |b_i|$. So in particular

$$ |S - S_N| = | \sum_{N}^{\infty} c_i | \leq \sum_{N}^\infty |b_i| \leq \epsilon/4 $$

We also have

$$ |S_N - S_{j,N} | < \epsilon / 4, \quad j > M $$

(where $M$ implicitly depends on $N$, and hence on $\epsilon$) and

$$ |S_j - S_{j,N} | < \epsilon / 4 $$

So we have that, applying the triangle inequality, that for every fixed $\epsilon > 0$, we can pick $M$ sufficiently large such that

$$ |S_j - S| < \epsilon \qquad \forall j > M$$

Q.E.D.

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Just a little tip: You might want to change | \sum_{N}^{\infty} c_i | to \left\lvert \sum_N^\infty c_i \right\rvert –  kahen Dec 19 '10 at 16:36
    
This is an extremely minor quibble, but why $\frac{\epsilon}{4}$ instead of $\frac{\epsilon}{3}$? Isn't the final line something like $|S_j - S| \leq |S_j - S_{j, N}| + |S_{j, N} - S_N| + |S_N - S| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$? Again, extremely minor, but just curious; I may be overlooking/misunderstanding something. –  Alex Basson Dec 19 '10 at 16:55
    
@kahen: if I were actually typing notes and papers, I would use the \abs macro which is defined to be precisely that in my preamble. I don't feel particular compelled to do so on Math.SE, though. –  Willie Wong Dec 19 '10 at 17:54
1  
@Alex: that's because I wrote it quickly without doing a draft on a piece of paper first. I thought I needed 4 pieces in the final sum originally, but got too lazy to go back and change everything to match when I found out I only needed 3. ;p –  Willie Wong Dec 19 '10 at 17:56

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