How do i evaluate this sum :$\sum _{m=1}^{\infty } \sum _{k=1}^{\infty } \frac{m(-1)^m(-1)^k\log(m+k)}{(m+k)^3}$?

Question

How do I evaluate the following sum:

$$\sum _{m=1}^{\infty } \sum _{k=1}^{\infty } \frac{m(-1)^m(-1)^k\log(m+k)}{(m+k)^3}$$

Note I used many idea such as :Hochino's Idea and taylor expansion of

$\log(1+x)$ at $x=1$ where $x=\frac{k}{m}$ ,but those methods not work .

and also i tried to write $\log(m+k)$ as a power series but it became to me as a

triple series then it is very complicated for evaluation !!!

Thank you for any help

Answer 1

$$\matrix{\sum_{m=1}^\infty\sum_{k=1}^\infty m\frac{(-1)^{m+k}\log(m+k)}{(m+k)^3} &=& \sum_{m=1}^\infty\sum_{k=1}^\infty\frac{(m+k)}{2}\frac{(-1)^{m+k}\log(m+k)}{(m+k)^3} & (1)\\&=& \sum_{n=1}^\infty\sum_{m+k=n} \frac{(-1)^n\log(n)}{2n^2} &(2)\\&=& \sum_{n=1}^\infty \frac{(-1)^n\log(n)(n-1)}{2n^2} &(3)\\&=& \frac{1}{2}(\eta'(1)-\eta'(2)) & (4)\\&=& \color{red}{\frac{1}{2} \gamma \log(2)-\frac{\log^2(2)}{4}-\frac{\pi^2}{24}\log(2)-\frac{\zeta'(2)}{4}} & (5)} $$

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• $(1)$ As Lucian points out in the comments: exchanging the summation labels $(m,k)\to (k,m)$ we get $S = \sum_m\sum_k \frac{m(-1)^{m+k}\log(m+k)}{(m+k)^3} = \sum_k\sum_m \frac{k(-1)^{m+k}\log(m+k)}{(m+k)^3}$ so $S+S = \sum_m\sum_k (m+k)\frac{(-1)^{m+k}\log(m+k)}{(m+k)^3}$ so $S = \sum_m\sum_k \frac{(m+k)}{2}\frac{(-1)^{m+k}\log(m+k)}{(m+k)^3}$.
• $(2)$-$(3)$ A sum on the form $\sum_{m=1}^\infty\sum_{k=1}^\infty a_{m+k}$ can be written $\sum_{n=1}^\infty w_n a_n$ where $w_n = \sum_{m+k=n}$ counts how many times $n$ can be written as a sum $n = m+k$ with $m,k\in\mathbb{N}$. This can be done in the $n-1$ ways $(m,k)= \{ (1,n-1), (2,n-2), \ldots, (n-1,1)\}$.
• $(4)$ As AchilleHui points out in the comments we have $\frac{\log(n)}{n^s} = -\frac{d}{ds}\frac{1}{n^s}$ so $\sum_{n=1}^\infty \frac{(-1)^n\log(n)}{n^s} = \frac{d}{ds}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \equiv \eta'(s)$ where $\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s)$ is the Dirichlet eta function and $\zeta(s)$ the Riemann zeta function.
• $(5)$ $\eta'(1) = \sum_{k=1}^\infty\frac{(-1)^{k}\log(k)}{k} = \gamma\log(2)-\frac{\log^2(2)}{2}$ where $\gamma$ is the Euler-Mascheroni constant is derived in this answer and $$\eta'(2) = \frac{d}{ds}\left[(1-2^{1-s})\zeta(s)\right]_{s=2} = \frac{\zeta '(2)}{2}+\frac{\log(2)}{2}\zeta(2) = \frac{\pi^2}{12}\log(2) + \frac{\zeta'(2)}{2}$$ where we have used $\zeta(2) = \frac{\pi^2}{6}$, see this question for many different ways to show this.

Answer 2

With all of the great comments, one of which suggested that one of us post an answer, I've decided to proceed. So, here we go ...

Let $S$ be the series of interest given by

$$S=\sum_{m=1}^{\infty}\sum_{k=1}^{\infty}\frac{m(-1)^{m+k}\log(m+k)}{(m+k)^3}$$

Now exploiting symmetry, we can write $S$ as

$$S=\frac12 \sum_{m=1}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{m+k}\log(m+k)}{(m+k)^2}$$

Next, we make the substutution $k=p-m$ and change the order of summation to reveal

$$\begin{align} S&=\frac12\sum_{p=2}^{\infty}\sum_{m=1}^{p-1}\frac{(-1)^p \log p}{p^2}\\\\ &=\frac12 \sum_{p=1}^{\infty}\frac{(-1)^p\log p}{p}-\frac12 \sum_{p=1}^{\infty}\frac{(-1)^p\log p}{p^2}\\\\ &=\frac12 \eta'(1)-\frac12 \eta'(2) \end{align}$$

where $\eta'(z)$ is derivative of the Dirichlet eta function. The Diriclet eta function is related to the Riemann zeta function by the expression

$$\eta(z)=(1-2^{1-z})\zeta(z)$$

The derivative $\eta'(z)$ can be written

$$\eta'(z)= \begin{cases} (1-2^{1-z})\zeta'(z)+2^{1-z}\log(2)\zeta(2)&, z>1\\\\ \left(\gamma-\frac12 \log(2)\right)\log(2)&,z=1 \end{cases}$$

Putting it all together gives

$$S=\frac12 \log(2)\left(\gamma-\frac12 \log(2)-\frac{\pi^2}{12}\right)-\frac14 \zeta'(2)$$