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I would like to compute:

$$ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \frac{(-1)^{i+j}}{i+j}$$

I wanted to use Fubini's theorem for double series but $$ \frac{(-1)^{i+j}}{i+j}_{(i,j)\in\mathbb{N*^2}}$$ is not a summable family for $$ \forall i>0$$

$$ \sum_{j=1}^{\infty} \frac{1}{i+j}=\infty$$

What am I supposed to do?

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  • $\begingroup$ Perhaps it would be useful to substitute $u=i+j$, so that the sum may be rewritten as $2\sum_{u=2}^\infty \frac{(-1)^u}{u}$ which converges by the alternating series test. $\endgroup$
    – nullUser
    Apr 11, 2012 at 21:07
  • $\begingroup$ @nullUser Except that there are $u-1$ couples $(i,j)$ such that $i+j=u$... $\endgroup$
    – Did
    Apr 11, 2012 at 21:11
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    $\begingroup$ @Matt In general, you can't re-arrange or re-combine the terms of a series that is not absolutely convergent. $\endgroup$ Apr 11, 2012 at 21:23
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    $\begingroup$ (We can, however, rearrange or alter a finite number of terms in a conditionally convergent series and obtain the expected answer.) $\endgroup$
    – anon
    Apr 11, 2012 at 21:46
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    $\begingroup$ @anon: +1, you beat me to it, while I was getting the link. $\endgroup$
    – bgins
    Apr 11, 2012 at 21:48

2 Answers 2

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How about: $$ -\int_0^1 (-x)^{i+j-1}\,dx = \frac{(-1)^{i+j}}{i+j} $$ then for each $x \in (0,1)$ we have $$ \sum_{i=1}^\infty\sum_{j=1}^\infty -(-x)^{i+j-1} = \frac{x}{(x+1)^2} $$ and integrate $$ \int_0^1\frac{x}{(x+1)^2}\,dx = \log 2 - \frac{1}{2} \approx 0.193147 $$

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Explanation for summation inside integral ... Two uses of this nice "monotone alternating" convergence theorem: Suppose $f_1(x) \ge 0\;$ is integrable on $E$ and $f_n(x) \searrow 0$ for almost every $x \in E$. Then $$ \sum_{n=1}^\infty (-1)^n \int_E f_n(x)\,dx = \int_E \left(\sum_{n=1}^\infty (-1)^n f_n(x)\right)\,dx $$ PROOF: Group the terms in pairs.

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More details now ... $$ \int_0^1 -(-x)^{i+j-1}\,dx = \frac{(-1)^{i+j}}{i+j} $$ For fixed $i$, the integrand decreases pointwise a.e. to zero in absolute value, and alternates sign. Therefore $$ \int_0^1 \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx = \sum_{j=1}^\infty\int_0^1 -(-x)^{i+j-1}\,dx = \sum_{j=1}^\infty \frac{(-1)^{i+j}}{i+j} $$ Now this integrand is $$ \sum_{j=1}^\infty-(-x)^{i+j-1} = \frac{-(-x)^i}{x+1} $$ As $i$ varies, this decreases a.e. to zero in absolute value, and alternates sign, so $$ \int_0^1 \sum_{i=1}^\infty \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx= \sum_{i=1}^\infty \int_0^1 \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx= \sum_{i=1}^\infty \sum_{j=1}^\infty \frac{(-1)^{i+j}}{i+j} $$

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  • $\begingroup$ Nice solution ! $\endgroup$ Apr 11, 2012 at 21:40
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    $\begingroup$ Which is in need of an explanation why the sum signs and the integral can be exchanged. $\endgroup$
    – Did
    Apr 11, 2012 at 21:46
  • $\begingroup$ The term by term integration theorem or the dominated convergence theorem can be used when there is only one parameter $n\in \mathbb{N}$. What about double integrals? $\endgroup$
    – Chon
    Apr 11, 2012 at 21:58
  • $\begingroup$ @GEdgar: Your answer would be enhanced if you wrote down the functions $f_n$ used in each of the two applications of the MACT. $\endgroup$
    – Did
    Apr 12, 2012 at 9:04
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    $\begingroup$ @Sasha: the original problem shows what I took to be an iterated sum... not a double sum... That is, it should be interpreted as $$\lim_{n\to\infty}\sum_{i=1}^n\;\lim_{m\to\infty}\sum_{j=1}^m \frac{(-1)^{i+j}}{i+j}$$ But of course it is not absolutely convergent, so summing in some other order may yield a different answer. You have shown some of those. So we conclude the "double sum" does not converge, although the "iterated sum" does. $\endgroup$
    – GEdgar
    Apr 13, 2012 at 12:35
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Let $S_n = \sum\limits_{i \geqslant 1, j \geqslant 1} \left( \frac{(-1)^{i+j}}{i+j} \mathbf{1}_{i+j \leqslant n+1} \right)$. Then $$ S_n = \sum_{i=1}^n \sum_{j=1}^{n+1-i} \frac{(-1)^{i+j}}{i+j} = \sum_{i=1}^n \sum_{j=1}^{n+1-i} \sum_{m=2}^{n+1} \delta_{i+j,m}\frac{(-1)^{i+j}}{i+j} = \sum_{m=2}^{n+1} \sum_{i=1}^n \sum_{j=1}^{n+1-i} \delta_{i+j,m}\frac{(-1)^{m}}{m} = \sum_{m=2}^{n+1} (-1)^m\frac{m-1}{m} = \sum_{m=2}^{n+1} (-1)^m - \sum_{m=2}^{n+1} \frac{(-1)^m}{m} = \frac{1}{2}\left( 1 - (-1)^n \right) + \sum_{m=1}^n \frac{(-1)^m}{m+1} $$

Notice that $$ \lim_{n \to \infty} S_{2n} = \lim_{n \to \infty} \sum_{m=1}^{2n} \frac{(-1)^m}{m+1} = \lim_{n \to \infty} \left( \ln(2) - 1 + H_{n+1/2} - H_n \right) = \ln(2) - 1 $$ and $$ \lim_{n \to \infty} S_{2n+1} = \lim_{n \to \infty} \left( 1 + \sum_{m=1}^{2n+1} \frac{(-1)^m}{m+1} \right) = \lim_{n \to \infty} \left( 1 + \ln(2) - 1 +H_{n+1/2} - H_{n+1} \right) = \ln(2) $$ Thus the sequence $S_n$ does not converges as $n$ increases, meaning that the original sum is not defined. The sequence $S_n$ has Cesaro mean, though: $$ \lim_{n \to \infty} \frac{1}{n} \sum_{m=1}^n S_m = \frac{1}{2} \left( \ln(2) + (\ln(2)-1)\right) = \ln(2) - \frac{1}{2} $$

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  • $\begingroup$ That's a brilliant reasoning ugh !!! A bit weird but brilliant haha :) $\endgroup$
    – ParaH2
    Jun 24, 2016 at 23:30

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