# Compute $\sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} \sum\limits_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$

How would you compute this sum? It's not a problem I need to immediately solve, but a problem that came to my mind today. I think that the generalization to more than three nested sums would be interesting as well.

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \sum_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$$

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By generalization, do you want $$\sum_{k_1,\dots,k_N=1}^{+\infty}\frac{(-1)^{\sum_{j=1}^Nk_j}}{\sum_{j=1}^Nk_j}‌​?$$ We can express the summand as an integral. – Davide Giraudo Jul 25 '12 at 10:13
Can you show us, why you think that it converges at all? – draks ... Jul 25 '12 at 10:31
Mathematica 8.0 is saying that the sum diverges. – James Fennell Jul 25 '12 at 10:44
@Chris'sister can you show/link your WA? – draks ... Jul 25 '12 at 10:49
@Chris'sister: in general it is better to use the MarkDown construction to post links. That is [link text](url to link to). I edited your comments above. This is one way to help prevent link breakage. – Willie Wong Jul 25 '12 at 15:58

Here is a simple lemma:

Let $(u_n)_{n\geqslant1}$ denote a decreasing sequence of positive functions defined on $(0,1)$, which converges pointwise to zero and such that $u_1$ is integrable on $(0,1)$. Then, $$\sum\limits_{n=1}^{+\infty}(-1)^n\int_0^1u_n(s)\,\mathrm ds=\int_0^1u(s)\,\mathrm ds,\qquad u(s)=\sum\limits_{n=1}^{+\infty}(-1)^nu_n(s).$$

Now, let us consider the multiple series the OP is interested in. One sees readily that it does not converge absolutely hence the idea is to apply the lemma three times.

• First, fix $n$ and $m$ and, for every $p\geqslant1$, consider $u_p(s)=s^{m+n+p-1}$. Then $u(s)=-\dfrac{s^{m+n}}{1+s}$ hence the lemma yields $$\sum\limits_{p=1}^{+\infty}\frac{(-1)^{m+n+p}}{m+n+p}=(-1)^{m+n}\sum\limits_{p=1}^{+\infty}(-1)^{p}\int_0^1u_p(s)\,\mathrm ds=(-1)^{m+n+1}\int_0^1\frac{s^{m+n}}{1+s}\,\mathrm ds.$$
• Second, fix $m$ and, for every $n\geqslant1$, consider $u_n(s)=\dfrac{s^{m+n}}{1+s}$. Then $u(s)=-\dfrac{s^{m+1}}{(1+s)^2}$ hence the lemma yields $$\sum\limits_{n=1}^{+\infty}(-1)^{m+n+1}\int_0^1\frac{s^{m+n}}{1+s}\,\mathrm ds=(-1)^m\int_0^1\frac{s^{m+1}}{(1+s)^2}\,\mathrm ds$$
• Third and finally, for every $m\geqslant1$, consider $u_m(s)=\dfrac{s^{m+1}}{(1+s)^2}$. Then $u(s)=-\dfrac{s^{2}}{(1+s)^3}$ hence the lemma yields $$\sum\limits_{m=1}^{+\infty}(-1)^m\int_0^1\frac{s^{m+1}}{(1+s)^2}\,\mathrm ds=-\int_0^1\frac{s^{2}}{(1+s)^3}\,\mathrm ds.$$

Thus, the triple series the OP is interested in converges and the value $S_3$ of the sum is $$\color{red}{S_3=-\int_0^1\frac{s^{2}}{(1+s)^3}\,\mathrm ds}=-\int_1^2\frac{s^{2}-2s+1}{s^3}\,\mathrm ds=-\left[\log(s)+\frac2s-\frac1{2s^2}\right]_1^2,$$ that is, $\color{red}{S_3=-\log(2)+\frac58}=-0.06814718\ldots$

The technique above shows more generally that, for every $k\geqslant1$, the analogous series over $k$ indices converges and that the value of its sum is $$S_k=(-1)^k\int_0^1\frac{s^{k-1}}{(1+s)^k}\,\mathrm ds=(-1)^k\left(\log(2)+\sum_{i=1}^{k-1}(-1)^i{k-1\choose i}\frac1i(1-2^{-i})\right).$$

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This is really, really nice. The $k=1$ case serves as another proof of the well-known alternating harmonic series sum,$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = \log(2)$$ – James Fennell Jul 25 '12 at 12:13
In fact, the structure of the general answer is quite interesting: a transcendental constant plus purely algebraic terms. – James Fennell Jul 25 '12 at 12:14
Chris' sister: The answer to this (where does this proof come from?) is in @James' remark: the method is an extension of a classical approach to prove that $S_1=-\log2$. – Did Jul 25 '12 at 12:21

This didn' fit in a comment $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left(\sum_{p=1}^{\infty}\frac{(-1)^{(m+n)+p}}{(m+n)+p}\pm(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}\right)\\ =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left((-1)^{(m+n)}\sum_{p=1}^{\infty}\frac{(-1)^{p}}{(m+n)+p}\pm(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}\right)\\ =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\left( \color{red}{ (-1)^{(m+n)}\sum_{p=1}^{\infty}\frac{(-1)^{p}}{(m+n)+p}+(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}}-(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}\right)\\ =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \color{red}{ (-1)^{(m+n)}\sum_{p=1}^{\infty}\frac{(-1)^{p}}{p}}-(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}\right)\\ =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \color{red}{ (-1)^{(m+n)}\log(2)}-\Phi_{\text{Lerch}}(-1, 1, 1+n+m)+(-1)^{m+n}\log(2)\\ =\underbrace{\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{(m+n)}2\log(2)}_{=0?}-\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \Phi_{\text{Lerch}}(-1, 1, 1+n+m)\\ =-\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \Phi_{\text{Lerch}}(-1, 1, 1+n+m)\;,$$ and this is where I give up for now. W|A can do some examples, that make me believe, that this doesn't converge...

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Each inner sum of the double sum underlined with $=0?$ diverges hence the double sum diverges. – Did Jul 25 '12 at 11:22
@did And why don't they cancel infinitely often to $0$? – draks ... Jul 25 '12 at 11:28
@did like Grandi's series... so let it be $0$ or $1$ or $1/2$. It would not affect to overall convergence/divergence... – draks ... Jul 25 '12 at 11:41
Well, if you have no qualm about summing $\sum\limits_n(-1)^n$, what can I say... – Did Jul 25 '12 at 11:54
@did maybe someting about: $\sum\limits_m\sum\limits_n (-1)^{m+n}=\sum\limits_m(-1)^{m}\sum\limits_n (-1)^{n}=\sum\limits_m(-1)^{m}G=G\sum\limits_m(-1)^{m}=G^2$, where $G$ is whatever you prefer as value for the Grandi series. So it converges, right? – draks ... Jul 25 '12 at 12:04