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I'm trying to find a closed form for the sum: $$\sum_{n=2}^\infty \frac{\ln(n)(-1)^n}{n^2} $$

It's been proven, $$\sum_{n=2}^\infty \frac{\ln(n)(-1)^n}{n} = \gamma\ln(2)-\frac{\ln(2)^2}{2}$$

Since they're similar, I feel there potentially could be a closed form solution of the first one. I want to know if there is also a closed form solution for the first summation.

I managed to rewrite the first sum as a double integral:

$$-\int_0^1\int_0^1 \frac{\ln(x)}{\ln(y)}\left(\frac{1}{1+x}-\frac{1}{1+xy} \right)dxdy $$ But I don't see any overt way of evaluating it. I've tried differentiating under the integral, but to no avail. I can't think of any change of variables that would help simplify as well.

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    $\begingroup$ Well, $\sum_{n=2}^\infty \frac{\ln(n)(-1)^n}{n} $ is the derivative of eta funtion in $1$, see here: en.wikipedia.org/wiki/Dirichlet_eta_function#Derivatives. Why not plug $s=2$? $\endgroup$
    – Zacky
    Commented Jan 4, 2019 at 15:22
  • $\begingroup$ I see thank you $\endgroup$
    – Tom Himler
    Commented Jan 4, 2019 at 15:26
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    $\begingroup$ You're welcome, but there is still some work to do. From here:en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant we can find a value for $\zeta'(2)$ and we should arrive at: $$\sum_{n=2}^\infty \frac{\ln(n)(-1)^n}{n}=\eta'(2)=\frac{\pi^2}{12}\left(\ln(4\pi) -12\ln(A) +\gamma\right)$$ $\endgroup$
    – Zacky
    Commented Jan 4, 2019 at 15:34
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    $\begingroup$ Just so you know, in your sum I think you forgot to square the denominator. Right now, the sum reads $\eta'(1)$. $\endgroup$
    – Diffusion
    Commented Jan 4, 2019 at 16:41
  • $\begingroup$ From differentiating $\eta(s)=(1-2^{1-s})\zeta(s)$, we obtain $\eta'(s)=\zeta(s)2^{1-s}\log 2+\big(1-2^{1-s}\big)\zeta'(s)$. However, I am not sure what happens near $s=1$, as the left side should be well defined, but the right side does not seem to be. $\endgroup$
    – Diffusion
    Commented Jan 4, 2019 at 22:35

1 Answer 1

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The definition of $\eta(s)$ is $$\eta(s)=(1-2^{1-s})\zeta(s)$$ Take the derivative, and we get by product rule $$\eta'(s)=2^{1-s}\ln(2)\zeta(s)+(1-2^{1-s})\zeta'(s)$$ $\eta(s)$ also has the following series representation $$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$$ Again, take the derivative $$\eta'(s)=\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^s}$$ Now, we can obtain a closed form $$\eta'(2)=\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^2}=2^{1-2}\ln(2)\zeta(2)+(1-2^{1-2})\zeta'(2)$$ Note that

$$\zeta'(2)=\frac{1}{6}\pi^2(-12\ln(A)+\ln(2\pi)+\gamma),~\zeta(2)=\frac{\pi^2}{6}$$

$$\begin{align} \sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^2}&=2^{-1}\ln(2)\zeta(2)+(1-2^{-1})\zeta'(2)\\ &=\frac{\pi^2}{12}\ln(2)+\frac{\pi^2}{12}(-12\ln(A)+\ln(2\pi)+\gamma)\\ &=\frac{\pi^2}{12}(\ln(4\pi)-12\ln(A)+\gamma) \end{align}$$

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  • $\begingroup$ very nice job indeed! $\endgroup$
    – G Cab
    Commented May 23, 2019 at 22:14
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    $\begingroup$ ($A$ is really defined as $e^{1/12-\zeta'(1)}$ appearing as the natural regularization of $\prod_{n=1}^\infty n^n$) $\endgroup$
    – reuns
    Commented May 23, 2019 at 22:34

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