TITLE: Compute $ \sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} \sum\limits_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$
QUESTION [14 upvotes]: 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.
$$ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \sum_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$$

REPLY [26 votes]: 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 
$$
S_k=(-1)^k\int_0^1\frac{s^{k-1}}{(1+s)^k}\,\mathrm ds=(-1)^k\left(\log(2)+\sum_{i=1}^{k-1}(-1)^i{k-1\choose i}\frac1i(1-2^{-i})\right).
$$