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I am trying to show $\sum\limits^\infty_{n=1} \left( {(-1)^n}\left( 1+n\ln(\frac{2n-1}{2n+1}) \right) \right)=\frac{2G}{\pi}-\frac{1}{2}$. I tried connecting this series to integrals in https://en.wikipedia.org/wiki/Catalan%27s_constant but was unable to link to any of them. I discovered the closed-form by using a piecewise function's Fourier series, however, I believe this is not a standard way. Do you have any ideas on this?

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    $\begingroup$ I am not sure if it helps but your sum is $$ \sum\limits_{n = 1}^\infty{( - 1)^n \left( {1 - 2n\tanh ^{ - 1} \left({\frac{1}{{2n}}} \right)} \right)} = \sum\limits_{n=1}^\infty {(-1)^n \sum\limits_{k = 1}^\infty {\frac{{-1}}{{(2k+1)(2n)^{2k} }}} } \\ = \sum\limits_{k = 1}^\infty {\frac{1}{{(2k + 1)2^{2k} }}\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n - 1} }}{{n^{2k} }}} } =\sum\limits_{k=1}^\infty {\frac{{(1-2^{1-2k} )\zeta(2k)}}{{(2k+1)2^{2k} }}} \\= \sum\limits_{k = 1}^\infty {\frac{{\zeta (2k)}}{{(2k + 1)2^{2k} }}}-2\sum\limits_{k=1}^\infty{\frac{{\zeta(2k)}}{{(2k+1)2^{4k} }}}. $$ $\endgroup$
    – Gary
    Commented Jan 4, 2022 at 7:49
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    $\begingroup$ You can write the sums as integrals using the Laurent series of the cotangent, thus your sum is $$ \int_0^{1/2} {(1 - \pi s\cot (\pi s))ds} - 4\int_0^{1/4} {(1 - \pi s\cot (\pi s))ds} \\ = - \int_0^{1/2} {\pi s\cot (\pi s)} + 4\int_0^{1/4} {\pi s\cot (\pi s)ds} - \frac{1}{2} \\ = - \frac{{\log 2}}{2} + \frac{{4G + \pi \log 2}}{{2\pi }} - \frac{1}{2} = \frac{{2G}}{\pi } - \frac{1}{2}. $$ $\endgroup$
    – Gary
    Commented Jan 4, 2022 at 8:12
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    $\begingroup$ These are all quite beneficial! $\endgroup$
    – mike
    Commented Jan 5, 2022 at 2:22

2 Answers 2

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@Gary has already provided the anwer to the problem via the integral. We can also find the solution in a different form, though it would be a heuristic (not rigorous) approach. Using the Frullani theorem, we can write the sum in the form

$$S=\sum_{n=1}^\infty (-1)^n n\bigg(\int_0^\infty e^{-nx}dx-\int_0^\infty \frac{e^{-nx}(e^{\frac{x}{2}}-e^{-\frac{x}{2}})}{x}dx\bigg)$$ $$=\sum_{n=1}^\infty(-1)^n\int_0^\infty n\,e^{-nx}\Big(1-\frac{(e^\frac{x}{2}-e^{-\frac{x}{2}})}{x}\Big)dx$$ We see that the integrand is well-defined at $x=0$ (equal to zeto at any $n$). Changing the order of summation and integration, we define $$S_1=\sum_{n=1}^\infty (-1)^nn\,e^{-nx}=-\frac{d}{dx}\sum_{n=1}^\infty (-1)^n\,e^{-nx}=\frac{d}{dx}\frac{1}{e^x+1}=-\frac{e^x}{(e^x+1)^2}$$ and the initial sum is $$S=-\int_0^\infty\frac{e^x}{(e^x+1)^2}dx+2\int_0^\infty\frac{e^x}{(e^x+1)^2}\frac{\sinh\frac{x}{2}}{x}dx=-\frac{1}{2}+\frac{1}{4}\int_{-\infty}^\infty\frac{\sinh t}{\cosh^2t}\frac{dt}{t}$$ The second integral can be evaluated, for example, via the integration in the complex plane (closing the contour by a big half-circle of radius $R$ in the upper half-plane). Integrand declines rapidly enough, so the integral along this half-circle $\to0$ as $R\to\infty$. $$\int_{-\infty}^\infty\frac{\sinh t}{\cosh^2t}\frac{dt}{t}=\oint=2\pi i \operatorname{Res}_{z=\frac{\pi i}{2}+2\pi i k}\frac{\sinh z}{\cosh^2z}\frac{1}{z}\,,\, k=0,1,2...$$ $$\sinh\Big(\frac{\pi i}{2}+2\pi ik+\epsilon\Big)=i(-1)^k\Big(1+O(\epsilon^2)\Big)$$ $$\cosh\Big(\frac{\pi i}{2}+2\pi ik+\epsilon\Big)=i(-1)^k\Big(\epsilon+O(\epsilon^3)\Big)$$ $$\text{at}\, z=\frac{\pi i}{2}+2\pi i k\qquad\frac{1}{z}=\frac{1}{\frac{\pi i}{2}+2\pi i k}\Big(1-\frac{\epsilon}{\frac{\pi i}{2}+2\pi i k}+O(\epsilon^2)\Big)$$ Taking all together and evaluating the residues $$\int_{-\infty}^\infty\frac{\sinh t}{\cosh^2t}\frac{dt}{t}=\sum_{k=0}^\infty\frac{2\pi}{\Big(\frac{\pi}{2}\Big)^2}\frac{(-1)^k}{(1+2k)^2}$$ $$S=-\frac{1}{2}+\frac{2}{\pi}\beta(2)=-\frac{1}{2}+\frac{2 G}{\pi}$$

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  • $\begingroup$ Wonderful! smart idea! $\endgroup$
    – mike
    Commented Jan 5, 2022 at 2:24
  • $\begingroup$ Nice solution! Minor point: you are missing a summation in the second line of the first displayed formula. $\endgroup$
    – Gary
    Commented Jan 5, 2022 at 3:20
  • $\begingroup$ @mike, thank you! $\endgroup$
    – Svyatoslav
    Commented Jan 5, 2022 at 7:55
  • $\begingroup$ @Gary, thank you. Correcting ) $\endgroup$
    – Svyatoslav
    Commented Jan 5, 2022 at 7:56
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Alternatively, your sum is equal to $$\begin{align}\sum_{n=1}^{\infty} (-1)^n \left(1+n\ln\left(\frac{2n-1}{2n+1}\right)\right)&=-\sum_{n=1}^{\infty}(-1)^n \left(n \ln\left(\frac{2n+1}{2n-1}\right)-1\right)\\&=-\sum_{n=1}^{\infty} (-1)^n (2n\operatorname{arccoth} (2n)-1)\end{align}$$

which I show how to derive and determine generalised results in my question on $$\sum_{n=1}^{\infty} (-1)^n (mn \operatorname{arccoth}(mn)-1)$$ by using the Mittag-Leffler pole expansion of $\csc (z)$ namely: $$\csc(z) = \frac{1}{z} + 2z \sum_{n=1}^{\infty} \frac{(-1)^n}{z^2 - \left(\pi n\right)^2}$$ and substituting it into the integral $$2G=\int_{0}^{\frac{\pi}{2}} x \csc x\, dx$$

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