<s>[INST] Sum involving $\ln{(2)}$ I got this sum. How do can this sum be equal to $8\ln{(2)}?$
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{35n-37}{(2n-1)(n-1)^2}+\frac{35n+37}{(2n+1)(n+1)^2}\right]=8\ln{(2)}$$
I have try to expand out the sum but it is too messy. Dealing the sum in this form, I haven't got any idea.
Any help.
 [/INST] First note that
\begin{eqnarray*} 
&&\frac{1}{n}\left[\frac{35n-37}{(2n-1)(n-1)^2}+\frac{35n+37}{(2n+1)(n+1)^2}\right]\\
&=&\frac{74}{n}+\frac2{(n+1)^2}-\frac2{(n-1)^2}+\frac{41}{n+1}+\frac{41}{n-1}-\frac{156}{2n+1}-\frac{156}{2n-1}
\end{eqnarray*}
and
\begin{eqnarray*}
&&\sum_{n=2}^{\infty}\frac{(-1)^n}{n-1}=\ln2,\sum_{n=2}^{\infty}\frac{(-1)^n}{n}=1-\ln2,\sum_{n=2}^{\infty}\frac{(-1)^n}{n+1}=-\frac12+\ln2,\\
&&\sum_{n=2}^{\infty}\frac{(-1)^n}{2n-1}=\frac{4-\pi}{4},\sum_{n=2}^{\infty}\frac{(-1)^n}{2n+1}=\frac{-8+3\pi}{12},\\
&&\sum_{n=2}^{\infty}(-1)^n\bigg[\frac2{(n+1)^2}-\frac2{(n-1)^2}\bigg]\\
&=&2\sum_{n=1}^{\infty}\bigg[\frac1{(2n+1)^2}-\frac1{(2n-1)^2}\bigg]-2\sum_{n=2}^{\infty}\bigg[\frac1{(2n)^2}-\frac1{(2n-2)^2}\bigg]\\
&=&-2+\frac12=-\frac32.
\end{eqnarray*}
Then you can put them to give the answer.
</s>