Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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. –  nullUser Apr 11 '12 at 21:07
@nullUser Except that there are $u-1$ couples $(i,j)$ such that $i+j=u$... –  Did Apr 11 '12 at 21:11
@Matt In general, you can't re-arrange or re-combine the terms of a series that is not absolutely convergent. –  Thomas Andrews Apr 11 '12 at 21:23
(We can, however, rearrange or alter a finite number of terms in a conditionally convergent series and obtain the expected answer.) –  anon Apr 11 '12 at 21:46
@anon: +1, you beat me to it, while I was getting the link. –  bgins Apr 11 '12 at 21:48

2 Answers 2

up vote 30 down vote accepted

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 $$ \int_0^1\frac{x}{(x+1)^2}\,dx = \log 2 - \frac{1}{2} \approx 0.193147 $$


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.


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 $$ \int_0^1 \sum_{i=1}^\infty \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx= \sum_{i=1}^\infty \int_0^1 \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx= \sum_{i=1}^\infty \sum_{j=1}^\infty \frac{(-1)^{i+j}}{i+j} $$

share|improve this answer
Nice solution ! –  Raymond Manzoni Apr 11 '12 at 21:40
Which is in need of an explanation why the sum signs and the integral can be exchanged. –  Did Apr 11 '12 at 21:46
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? –  Chon Apr 11 '12 at 21:58
@GEdgar: Your answer would be enhanced if you wrote down the functions $f_n$ used in each of the two applications of the MACT. –  Did Apr 12 '12 at 9:04
@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. –  GEdgar Apr 13 '12 at 12:35

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: $$ \lim_{n \to \infty} \frac{1}{n} \sum_{m=1}^n S_m = \frac{1}{2} \left( \ln(2) + (\ln(2)-1)\right) = \ln(2) - \frac{1}{2} $$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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