Closed form of $\sum\limits^\infty_{n=1} \left( {(-1)^n}\left( 1+n\ln(\frac{2n-1}{2n+1}) \right) \right)$ 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?
 A: @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
$$\int_{-\infty}^\infty\frac{\sinh t}{\cosh^2t}\frac{dt}{t}=\sum_{k=0}^\infty\frac{2\pi}{\Big(\frac{\pi}{2}\Big)^2}\frac{(-1)^k}{(1+2k)^2}$$
$$S=-\frac{1}{2}+\frac{2}{\pi}\beta(2)=-\frac{1}{2}+\frac{2 G}{\pi}$$
A: 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
$$2G=\int_{0}^{\frac{\pi}{2}} x \csc x\, dx$$
