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$$ \sum_{n_1 , n_2 , \cdots , n_m \ge 1} \frac{(-1)^{n_1 + n_2 + \cdots + n_m}}{n_1 +n_2 +\cdots +n_m}.$$ What about if we change the denominator into $ {n_1}^k + {n_2}^k + \cdots + {n_m}^k $ for $k \ge 2$?

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This question does not show any research effort – The Chaz 2.0 Apr 25 '12 at 13:06
Note that if all those sums really start at zero then one term in the sum is $1/0$. – Gerry Myerson Apr 25 '12 at 13:07
@GerryMyerson: so the result is $\infty$ :-) – Fabian Apr 25 '12 at 13:12
No, @Fabian, $1/0$ is undefined, not $\infty.$ – Thomas Andrews Apr 25 '12 at 13:14
@hkju See this post which treats $m=2$ case. I hope you can extend to general $m$. – Sasha Apr 25 '12 at 13:21

First, of all, the multiple sum $$ \sum_{n_1 >0, n_2 >0, \ldots, n_m >0 } \frac{(-1)^{n_1+n_2+\cdots+n_m}}{n_1+n_2+\cdots+n_m} $$ does not converge. Indeed, consider regularized version $$ \begin{eqnarray} S_k &=& \sum_{n_1 >0, n_2 >0, \ldots, n_m >0 } \frac{(-1)^{n_1+n_2+\cdots+n_m}}{n_1+n_2+\cdots+n_m} I_{n_1+n_2+ \cdots + n_m \leqslant k} \\ &=& \sum_{p=m}^{k} \sum_{n_1 >0, n_2 >0, \ldots, n_m >0 } \frac{(-1)^{n_1+n_2+\cdots+n_m}}{n_1+n_2+\cdots+n_m} \delta_{n_1+n_2+\cdots+n_m,p} I_{n_1+n_2+ \cdots + n_m \leqslant k} \\ &=& \sum_{p=m}^k \frac{(-1)^p}{p} \sum_{n_1=1}^{p-m+1} \sum_{n_2=1}^{p-m+1} \cdots \sum_{n_m=1}^{p-m+1} \delta_{n_1+n_2+\cdots+n_m,p} \\ &=& \sum_{p=m}^k \frac{(-1)^p}{p} \sum_{n_1=0}^{p-m} \sum_{n_2=0}^{p-m} \cdots \sum_{n_m=0}^{p-m} \delta_{n_1+n_2+\cdots+n_m,p-m} \\ &=& \sum_{p=0}^{k-m} \frac{(-1)^{p+m}}{p+m} \sum_{n_1=0}^{p} \sum_{n_2=0}^{p} \cdots \sum_{n_m=0}^{p} \delta_{n_1+n_2+\cdots+n_m,p} \\ &=& \sum_{p=0}^{k-m} \frac{(-1)^{p+m}}{p+m} [z]^p \left( \frac{1-z^{p+1}}{1-z}\right)^m = \sum_{p=0}^{k-m} \frac{(-1)^{p+m}}{p+m} [z]^p \left( \frac{1}{1-z}\right)^m \\ &=& \sum_{p=0}^{k-m} \frac{(-1)^{p+m}}{p+m} \binom{m+p-1}{p} \end{eqnarray} $$ The sequence $S_k$ does not converge to any limit. In fact, for large $k$, $$ S_k = \frac{(-1)^m}{m} {}_2 F_1 \left(m,m;m+1,-1\right) + \frac{(-1)^k k^{m-2}}{2 (m-1)!} \left( 1+ \mathcal{o}(1)\right) $$

However, the sum can be regularized, see @G.Edgard's excellent answer discussing $m=2$ case.

The regularization amounts to exchanging the summation and integration here $$ \begin{eqnarray} S_\text{reg} &=& \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_m=1}^\infty (-1)^{n_1+n_2+\cdots+n_m} \int_0^1 x^{n_1+n_2+\cdots+n_m-1} \mathrm{d} x \\ &\stackrel{\color\maroon{\text{regularization}}}{=} & \int_0^1 \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_m=1}^\infty (-1)^{n_1+n_2+\cdots+n_m} x^{n_1+n_2+\cdots+n_m-1} \mathrm{d} x \\ &=& \int_0^1 \left( \sum_{n=1}^\infty (-1)^n x^n \right)^m \frac{\mathrm{d} x}{x} \\ &=& \int_0^1 \left( \frac{-x}{1+x} \right)^m \frac{\mathrm{d} x}{x} = (-1)^m \int_0^1 x^{m-1} \left(1+x\right)^{-m} \mathrm{d} x \\ &\stackrel{\color\maroon{\text{by parts}}}{=}& (-1)^m \left. \left( -\sum_{k=0}^{m-2} \frac{x^{m-1-k} (1+x)^{1-m+k}}{m-k-1} + \log(1+x) \right) \right|_{0}^1 \\ &=& (-1)^m \left( \log(2) - \sum_{k=0}^{m-2} \frac{2^{1+k-m}}{m-k-1} \right) \\&\stackrel{k \to m-1-k}{=} & (-1)^m \left( \log(2) - \sum_{k=1}^{m-1} \frac{2^{-k}}{k} \right) \end{eqnarray} $$

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This sum (starting at 1 as noted by Gerry Myerson) can be transformed into an integral by noting that $$ \frac1N=\int_0^\infty\!d\lambda\,e^{-\lambda N}.$$

With $N=\sum_{m=1}^m n_m$, we have (assuming everything is well-defined such that we can interchange summation and integration) $$\begin{align}\sum_{n_1 =1}^\infty \sum_{n_2 =1}^\infty \cdots \sum_{n_m =1}^\infty \frac{(-1)^{n_1 + n_2 + \cdots + n_m}}{n_1 +n_2 +\cdots +n_m} &= \int_0^\infty\!d\lambda \prod_{i=1}^m \overbrace{\sum_{n_i=1}^\infty (-1)^{n_i} e^{-\lambda n_i}}^{-1/(1+e^{\lambda})}\\ &=(-1)^m\int_0^\infty \!d\lambda\,(1+e^{\lambda})^{-m}\end{align}.$$

The integral remains to be evaluated. note that $$\begin{align}\int_0^\infty \!d\lambda\,(1+e^{\lambda})^{-m} &= \int_0^\infty \!d\lambda\, e^{-m\lambda} \overbrace{(1+e^{-\lambda})^{-m}}^{\sum_{k=0}^\infty \binom{m+k-1}{m-1} (-1)^k e^{-km}}\\ &= \sum_{k=0}^\infty \binom{m+k-1}{m-1} \frac{(-1)^k}{m+k}, \end{align} $$ which as far as I see cannot be expressed in terms of elementary functions.

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Moreover, such sums are divergent for $m \geqslant 2$. – Sasha Apr 25 '12 at 13:53
@Sasha: oh, you are right... – Fabian Apr 25 '12 at 13:57
See my post for the closed form result for your integral, stated at the end of the post. – Sasha Apr 25 '12 at 16:39

Define $f(x)=\displaystyle\sum_{n\geq 0}(-1)^nx^n$. The given expression is then $\int_0^1f(x)^mdx$.

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-1, the said integral does not reproduce the sum, rather $\int_0^1 f(x)^m \frac{dx}{x}$ does. – Sasha Apr 25 '12 at 15:05

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