<s>[INST] Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)+1\right)=4\ln\left(\!\frac{3}{4}\!\right)+1+9\ln\left(\!\frac{8}{9}\!\right)+1+\ldots$$
Any tricks to solve it?
 [/INST] This answer is just an elaboration on KStarGamer's comment. I myself went for the calculation idea.
Suppose the sum is converging. We have then
$$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$
$$\implies S'(x)=\sum_{n=2}^{\infty}\left(-\frac{2x}{n^2}n^2\frac{1}{1-\frac{x^2}{n^2}}+2x\right)$$
\begin{align*}
S'(x)&=\sum_{n=2}^{\infty}\left(-2x{\frac{n^2}{n^2-x^2}}+2x\right)\\
&=2x\sum_{n=2}^{\infty}\left({\frac{n^2}{x^2-n^2}}+1\right)\\
&=2x\sum_{n=2}^{\infty}\left({\frac{n^2+x^2-n^2}{x^2-n^2}}\right)\\
&=x^2\left[2x \sum_{n=2}^{\infty}{\frac{1}{x^2-n^2}}\right]\\
&=\pi x^2\left[2\pi x \sum_{n=2}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}\right]\\
&=\pi x^2\left[-\frac{2x}{\pi(x^2-1)}-\frac{1}{\pi x}+\underbrace{\frac{1}{\pi x}+2\pi x \sum_{n=1}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}}_{\cot(\pi x)}{}\right]\\
\end{align*}
where $\cot(\pi x)$ is expressed using Mittag-Leffler's theorem on meromorphic functions (Look here).
We finally get that
$$S'(x)=-x+\left(\pi x^2\cot(\pi x)-\frac{2x^3}{x^2-1}\right)$$
By integrating, we get :
$$S(x)=-\frac{x^2}2+\int_0^x\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt$$
For $x=1$, we get
$$S(1)=-\frac{1}{2}+\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt=-\frac{1}{2}+\ln(\pi)-1=\ln(\pi)-\frac{3}{2}.$$
The integral $\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt$ was evaluated exactly via Wolframalpha.
Thanks to both Angelo and KStarGamer for their corrections and comments.
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