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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

Salmahamizi Hamizi
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  • I did this but what about m ? – Salmahamizi Hamizi Dec 25 '15 at 00:32
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    The sum equals to $\left(\gamma - \frac{\log(2)}{2} - \frac{\pi^2}{12}\right)\log(2) - \frac{\zeta'(2)}{2} \approx 0.0585523255789$ where $\zeta(s)$ is the Riemann zeta function. The key is $n^{-s}\log n = -\frac{d}{ds} n^{-s}$ – achille hui Dec 25 '15 at 00:40
  • did you meant van mangoldt function ? – Salmahamizi Hamizi Dec 25 '15 at 00:41
  • @achillehui I think you forgot a factor of $1/2$ in front of that (whole) expression. At least that is what I get (and I checked numerically that the result is $0.0292747 = \frac{1}{2}\cdot 0.0585523$). – Winther Dec 25 '15 at 00:43
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    Hint: Due to symmetry, $$\sum_{m=1}^{\infty}\sum_{k=1}^{\infty}\frac{m(-1)^{m+k}\log(m+k)}{(m+k)^3} ~=~ \sum_{m=1}^{\infty}\sum_{k=1}^{\infty}\frac{k(-1)^{m+k}\log(m+k)}{(m+k)^3}$$ Now add the two together, and see what happens. – Lucian Dec 25 '15 at 00:46
  • Very clever stuff, but somebody should post them as answers they seem to go beyond comments. – Gregory Grant Dec 25 '15 at 00:47
  • Further Hint: Let $n=m+k.$ In how many ways can we write n as the sum of two positive integers ? – Lucian Dec 25 '15 at 00:55
  • @Winther oops, you are right. that expression is off by a factor $\frac12$. – achille hui Dec 25 '15 at 00:56
  • @Lucian,may you would like to use this series to know the number of way to write it as sum of 2 positive integer :$$ \sum_{n=1}^{\infty} \frac{R(n)}{n^s} = 4\zeta(s) L(s,\chi_{-4}), $$ where R(n) is the number of way – Salmahamizi Hamizi Dec 25 '15 at 01:02
  • @achillehui It might be better to elude to the Dirichlet eta function, which has, of course, a simple relationship to the Riemann zeta function. I'm suggesting this only because the series of interest is alternating and therefore the Dirichlet eta is more natural. Happy Holidays - Mark – Mark Viola Dec 25 '15 at 01:04

2 Answers2

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

  • $(1)$ 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.
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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

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

Mark Viola
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    in the line 2 in this :$$\begin{align} S&=\frac12\sum_{p=2}^{\infty}\sum_{m=1}^{p-1}\frac{(-1)^p \log p}{p^2}\\ &=\frac12 \sum_{p=1}^{\infty}\frac{(-1)^p\log p}{p^2}-\frac12 \sum_{p=1}^{\infty}\frac{(-1)^p\log p}{p^2}\\ &=\frac12 \eta'(1)-\frac12 \eta'(2) \end{align}$$ by your sum should be nule they are the same sum – Salmahamizi Hamizi Dec 25 '15 at 13:55
  • @SalmahamiziHamizi Thank you for the catch! +1 . The first sum in Line 2 has $p$, not $p^2$ in the denominator. I've edited. Happy Holidays! - Mark – Mark Viola Dec 25 '15 at 16:26
  • I edited before but it were rejected i don't know why ? – Salmahamizi Hamizi Dec 25 '15 at 16:32
  • @SalmahamiziHamizi I don't know why, but thank you for trying. - Mark – Mark Viola Dec 25 '15 at 16:37