Summing Lerch Transcendents The Lerch transcendent
is given by
$$
    \Phi(z, s, \alpha) = \sum_{n=0}^\infty \frac { z^n} {(n+\alpha)^s}.
$$
While computing  $\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}
\sum_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$,
the expression
$$
-\sum_{k=1}^{\infty} \Phi(-1, 1, 1+k)
$$
came up. Is there an (easy?) way to calculate that?
Writing it down, it gives:
$$
-\sum_{k=1}^{\infty} \Phi(-1, 1, 1+k)=
\sum_{k=1}^{\infty} \sum_{n=1}^\infty \frac { (-1)^n} {n^{k+1}}
$$
Is changing the summation order valid?
There is a relation to the Dirichlet $\eta$ function
$$
    \eta(s) = \sum_{n=1}^{\infty}{(-1)^{n-1} \over n^s} =
\frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots
$$
but (how) can I use that? The double series then reads
$$
\sum_{k=1}^{\infty} \sum_{n=1}^\infty \frac { (-1)^n} {n^{k+1}}=
-\sum_{k=1}^{\infty} \eta(k+1).
$$
Interestingly, that among the values for $\eta$, given at the WP, you'll find $\eta(0)=1/2$ related to Grandi's series and $\eta(1)=\ln(2)$, both show up in my attempt to prove the convergence of the triple product given there.  
 A: 
The double series $\displaystyle\sum_{k=1}^{\infty} \sum_{n=1}^\infty \frac { (-1)^n} {n^{k+1}}$ diverges.

To see this, one can imitate the strategy used in the answer to the other question, and use the identity
$$
\frac1{n^{k+1}}=\int_0^{+\infty}\mathrm e^{-ns}\frac{s^k}{k!}\,\mathrm ds.
$$
Thus, for every $k\geqslant1$,
$$
\sum_{n=1}^\infty \frac { (-1)^n} {n^{k+1}}=\int_0^{+\infty}\sum_{n=1}^\infty(-1)^n\mathrm e^{-ns}\frac{s^k}{k!}\,\mathrm ds=\int_0^{+\infty}\frac{-\mathrm e^{-s}}{1+\mathrm e^{-s}}\frac{s^k}{k!}\,\mathrm ds.
$$
Since $1+\mathrm e^{-s}\leqslant2$ uniformly on $s\geqslant0$,
$$
\sum_{n=1}^\infty \frac { (-1)^n} {n^{k+1}}\leqslant-\frac12\int_0^{+\infty}\mathrm e^{-s}\frac{s^k}{k!}\,\mathrm ds=-\frac12.
$$
This proves that the double series diverges.
Edit: More directly, each series $\displaystyle\sum_{n=1}^\infty \frac { (-1)^n} {n^{k+1}}$ is alternating hence it converges and the value of its sum is between any two successive partial sums. 
For example, $\displaystyle\sum_{n=1}^\infty \frac { (-1)^n} {n^{k+1}}\leqslant\sum_{n=1}^2 \frac { (-1)^n} {n^{k+1}}=-1+\frac1{2^{k+1}}\leqslant-\frac34$ for every $k\geqslant1$. QED.
A: With the definition of the Dirichlet's $\eta$ as in your post
$ \sum_{k=1}^\infty \eta(k+1) $
the whole expression is divergent because the $\eta(k+1)$ converge to $1$.        
But if you rewrite your formula
$$ \sum_{k=1}^\infty (\eta(k+1)-1) $$ this shall converge because $\eta(1+k)-1$ converges quickly to zero when $k$ increases. Finally we get for this "residuum" 
$$ \sum_{k=1}^\infty (\eta(k+1)-1) \to 1- \ln(4) $$ 
(I've guessed the $\ln(4)$ by empirical approximation but I'm sure the sum of $(\eta(k)-1)$ are known exactly (and to be $=-\log 2$) 
After this one can write down the limit
$$ \lim_{K\to\infty} \sum_{k=1}^K (\eta(k+1)) - K =  1- \ln(4) $$
