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.