Closed form for the sum of the series $\sum^\infty_{n=1} \left( {(-1)^n}\left( 1+n\ln(\frac{2n-1}{2n+1}) \right) \right)$ If $a>2$ then $\sum\limits^\infty_{n=1} \left( {(-1)^n}\left( 1+n\ln(\frac{an-1}{an+1}) \right) \right)$ diverges by divergent test. Does it converge if $a=2$? Is it possible to find an exact form for the sum of the series in case it converges?

One has
\begin{align*} {(-1)^n}\left( 1+n\ln\left(\frac{2n-1}{2n+1}\right) \right) &= {(-1)^n}\left( 1+n\ln\left(1 -\frac{2}{2n+1}\right) \right)\\
&= {(-1)^n}\left( 1+n \left(- \frac{2}{2n+1} - \frac{2}{(2n+1)^2}  + O \left( \frac{1}{n^3}\right)\right) \right)\\
&= \frac{(-1)^n}{2n+1}\left( \frac{1}{2n+1}  + O \left( \frac{1}{n}\right)\right) \\
&= \frac{(-1)^n}{(2n+1)^2} + O \left( \frac{1}{n^2}\right) \\
\end{align*}
which proves the (absolute) convergence of the series for $a=2$.
