<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] Here's an elementary proof,  assuming that we're allowed to use Stirling's formula:
$$\ln (n!) = \sum_{k=2}^n \ln(k) = n\ln n -n + \frac 1 2 \ln n + \frac 1 2 \ln 2\pi + \mathcal O(\frac 1 n)$$
Let us study the partial sums, denoting $u_k=k^2 \ln \left(1-\frac 1 {k^2} \right) + 1$:
$$\begin{align}
\sum_{k=2}^n u_k &=\sum_{k=2}^n \left(k^2 \ln \left(1-\frac 1 {k^2} \right) + 1 \right) \\
&=\sum_{k=2}^n \left(k^2\ln \left(\frac {(k+1)(k-1)} {k^2}\right) + 1 \right) \\
&=\sum_{k=2}^n \left(k^2 \ln(k+1) + k^2\ln(k-1) - 2k^2\ln k + 1 \right) \\
&=-\ln 2 + n^2\ln(n+1) - (n+1)^2\ln n +\sum_{k=2}^n \left((k-1)^2 \ln k + (k+1)^2\ln k - 2k^2\ln k + 1 \right) \\
&=-\ln 2 + n^2\ln(n+1) - (n+1)^2\ln n +\sum_{k=2}^n \left(2\ln k + 1 \right) \\
&=-\ln 2 + \left(n^2\ln n + n^2 \ln(1+\frac 1 n)\right) - \left(n^2\ln n + 2n\ln n + \ln n\right) \\ &\ \ \ \ +\left(2n\ln n - 2n + \ln n + \ln 2\pi + \mathcal O(\frac 1 n)\right) + (n-1) \\
&=-\ln 2 + n^2 \ln(1+\frac 1 n) - n - 1  + \ln 2\pi + \mathcal O(\frac 1 n) \\
&=n^2 \left(\frac 1 n - \frac 1 {2n^2} + O(\frac 1 {n^3})\right) - n - 1  + \ln\pi + \mathcal O(\frac 1 n) \\
&=- \frac 3 2 + \ln\pi + \mathcal O(\frac 1 n)
\end{align}$$
Therefore, as desired: $$\sum_{k=2}^{+\infty} u_k = \ln\pi - \frac 3 2$$
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