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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.

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If the series converges absolutely, then you can change the order of summation – Mhenni Benghorbal Jul 25 '12 at 12:18
@MhenniBenghorbal $\sum_{n=1}^\infty \frac { 1} {n^{k+1}}=\zeta(k+1)$, so it converges absolutely for $k\ge 1$, right? – draks ... Jul 25 '12 at 12:45
draks:The double series has to converge absolutely. – Mhenni Benghorbal Jul 25 '12 at 12:53
@MhenniBenghorbal and the double doesn't: $\lim_{n\to \infty}\sum_2^n \zeta(n)=\infty $. Thanks – draks ... Jul 25 '12 at 13:02
@draks:any time. – Mhenni Benghorbal Jul 25 '12 at 13:06
up vote 3 down vote accepted

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.

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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 $$ \lim_{K\to\infty} \sum_{k=1}^K (\eta(k+1)) - K = 1- \ln(4) $$

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