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.
$$ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \sum_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$$
 A: 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 
$$
S_k=(-1)^k\int_0^1\frac{s^{k-1}}{(1+s)^k}\,\mathrm ds=(-1)^k\left(\log(2)+\sum_{i=1}^{k-1}(-1)^i{k-1\choose i}\frac1i(1-2^{-i})\right).
$$
A: 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...
Ref's: $-(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}=-\Phi_{\text{Lerch}}(-1, 1, 1+n+m)+(-1)^{m+n}\log(2)$
