Find the sum $\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right)$ based on $a^2-b^2 $? Find the sum of the  given series
$$\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right).$$
My attempt :  I used the formula  $a^2- b^2 = (a+b)(a-b)$, then I get 
$$\sum_{n=1}^{\infty} \left((-1)^{n-1}\log \left(1- \frac{1}{n+1}\right) +(-1)^{n-1}\log \left(1+\frac{1}{n+1}\right)\right).$$
After that I am not able  to proceed further
Any hints/solution will be appreciated.
Thank you.
 A: Based on the link that Jack provided, 
$${\begin{aligned}
\zeta (s)&=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\Re (s)>1\\\eta (s)&=(1-2^{1-s})\zeta (s)\\
&=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}},\Re (s)>0
\end{aligned}}$$
where $\zeta (s)$ is the zeta function, and $\eta (s)$ is the eta function. Then we can evaluate your series by using Euler transformation and taking the derivative of the eta function. You just need to know the chain rule and the exponential derivatives.
Using Euler Transform for the eta function, we have
$$\eta (s)={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {1}{n^{s}}}-{\frac {1}{(n+1)^{s}}}\right],\Re (s)>-1$$
Taking the derivative,
$$\begin{aligned}
\eta '(s)&=(1-2^{1-s})\zeta '(s)+2^{1-s}(\ln 2)\zeta (s)\\&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {\ln n}{n^{s}}}-{\frac {\ln(n+1)}{(n+1)^{s}}}\right],\Re (s)>-1
\end{aligned}$$
$$\begin{aligned}
\Rightarrow \eta '(0)&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[\ln n-\ln(n+1)\right]\\&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}\\&=-{\frac {1}{2}}\left(\ln {\frac {1}{2}}-\ln {\frac {2}{3}}+\ln {\frac {3}{4}}-\ln {\frac {4}{5}}+\ln {\frac {5}{6}}-\cdots \right)\\&={\frac {1}{2}}\left(\ln {\frac {2}{1}}+\ln {\frac {2}{3}}+\ln {\frac {4}{3}}+\ln {\frac {4}{5}}+\ln {\frac {6}{5}}+\cdots \right)\\&={\frac {1}{2}}\ln \left({\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot \cdots \right)
\end{aligned}$$
Recall Wallis' product,
$$\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {\pi }{2}}$$
So
$$\eta '(0)=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}=\frac{1}{2}\ln{\frac {\pi }{2}}\\
\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}=-\ln\frac{\pi}{2}$$
Using the same logic, we have
$$\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n+1}{n+2}}=-\ln\frac{\pi}{4}$$
Then,
$$\begin{align}
\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right)&=\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}+\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n+1}{n+2}}\\
&=-\ln\frac{\pi}{2}-\ln\frac{\pi}{4}\\
&=\color{red}{-\ln\frac{\pi^2}{8}}
\end{align}$$
A: Hint: Note that for all $n$ you have
$$\log\left(1-\frac{1}{(n+1)^2}\right)=\log\left(\frac{n(n+2)}{(n+1)^2}\right)=\log(n)-2\log(n+1)+\log(n+2).$$
A: i have an idea to supply Larry's proof:
$log(1-\frac{1}{n+1})+log(1+\frac{1}{n+1})=2[\frac{(1/(n+1))^2}{2}+\frac{(1/(n+1))^4}{4}+......]$
then rewrite above to $2*(1/2)*(t^2+t^4/2+t^6/3+......)$
now your question is improved to be $\sum{(-1)^{n}e^{(\frac{1}{n+1})^2}}=-e^{(1/2)^2}+e^{(1/3)^2}-e^{(1/4)^2}+......$
so this sum is equal to $-In(3/4)\zeta{(2)}$ since the first two terms should be eliminate after a transformation from eta function to zeta function!
