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?
 A: 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: $$\sum_{k=2}^{+\infty} u_k = \ln\pi - \frac 3 2$$
A: Just a little simplification (without usage of WolframAlpha) to the nice posted solutions.
Using the similar approach, denoting
$$f(x)=\sum^\infty_{n=2}n^2\ln\left(1-\frac{x}{n^2}\right)+x$$
and taking the derivative with respect to $x$:
$$f'(x)=\sum_{n=2}^\infty\frac{x}{x-n^2}=\sum_{n=1}^\infty\frac{x}{x-n^2}-\frac{x}{x-1}$$
Using $\,\,\displaystyle \pi\cot\pi z=\frac{1}{z}+2\sum_{n=1}^\infty\frac{z}{z^2-n^2}$
$$f'(x)=\frac{\pi}{2}\sqrt x\cot(\pi\sqrt x)-\frac{1}{2}-\frac{x}{x-1}=-\frac{3}{2}+\frac{\pi}{2}\sqrt x\cot(\pi\sqrt x)-\frac{1}{x-1}$$
and
$$f(1)=\int_0^1f'(x)dx=-\frac{3}{2}+\lim_{\epsilon\to 0}\int_0^{1-\epsilon}\bigg(\frac{\pi}{2}\sqrt x\cot(\pi\sqrt x)-\frac{1}{x-1}\bigg)dx$$
$$=-\frac{3}{2}+\lim_{\epsilon\to 0}\Big(I_1(\epsilon)+I_2(\epsilon)\Big)\tag{1}$$
where
$$I_2(\epsilon)=\int_0^{1-\epsilon}\frac{dx}{1-x}=-\ln\epsilon\tag{2}$$
and
$$I_1(\epsilon)=\frac{\pi}{2}\int_0^{1-\epsilon}\sqrt x\cot(\pi\sqrt x)\,dx=\frac{1}{\pi^2}\int_0^{\pi\sqrt{1-\epsilon}}\cot(t)t^2dt$$
Integrating by part,
$$I_1(\epsilon)=\frac{t^2\ln\sin t}{\pi^2}\,\bigg|_0^{\pi\sqrt{1-\epsilon}}-\frac{2}{\pi^2}\int_0^{\pi\sqrt{1-\epsilon}}\ln(\sin t)t dt$$
The second term converges at $\epsilon =0$, therefore
$$I_1(\epsilon)=\ln\sin(\pi\sqrt{1-\epsilon})-\frac{2}{\pi^2}\int_0^{\pi}\ln(\sin t)t dt+O(\epsilon)$$
$$=\ln\frac{\pi \epsilon}{2}-\frac{2}{\pi^2}\int_0^{\pi}\ln(\sin t)t dt+O(\epsilon)\tag{3}$$
where
$$\int_0^{\pi}\ln(\sin t)t dt=\int_0^{\frac{\pi}{2}}\ln(\sin t)tdt+\int_0^{\frac{\pi}{2}}\ln(\cos t)\Big(t+\frac{\pi}{2}\Big)dt$$
$$=\int_0^{\frac{\pi}{2}}\ln(\sin t)t dt+\int_0^{\frac{\pi}{2}}\ln(\sin t)\Big(\frac{\pi}{2}-t\Big) dt+\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\ln(\cos t)dt$$
Using $\displaystyle \int_0^{\frac{\pi}{2}}\ln(\cos t)dt=\int_0^{\frac{\pi}{2}}\ln(\sin t)dt=-\frac{\pi}{2}\ln 2$
$$\int_0^{\pi}\ln(\sin t)t dt=\pi\int_0^{\frac{\pi}{2}}\ln(\sin t)dt=-\frac{\pi^2}{2}\ln2$$
Using (3), we have for $I_1(\epsilon)$
$$I_1(\epsilon)=\ln\frac{\pi \epsilon}{2}+\ln2+O(\epsilon)\tag{4}$$
Putting (2) and (4) into (1),
$$f(1)=-\frac{3}{2}+\lim_{\epsilon\to 0}\Big(-\ln\epsilon+\ln\frac{\pi \epsilon}{2}+\ln2+O(\epsilon)\Big)=-\frac{3}{2}+\ln\pi$$
A: 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.
A: (I have partially copied @Hamdiken's answer to make it more complete)
Define the sum
$$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$
and differentiate with respect to $x$ to obtain
\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*}
by the Mittag-Leffler pole expansion of $\cot(\pi x)$.
We finally get that
$$S'(x)=-x+\left(\pi x^2\cot(\pi x)-\frac{2x^3}{x^2-1}\right)$$
Upon integrating, we determine
$$S(x)=-\frac{x^2}2+\int_0^x\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)\,dt$$
For $x=1$, we get
$$S(1)=-\frac{1}{2}+\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)\,dt\stackrel{t\to 1-t}{=}-\frac{3}{2}-\ln(2)+\int_{0}^{1} \left(\frac{1}{t} -\pi (1-t)^2\cot(\pi t)\right)\, dt
$$
Now by the Laurent series for $\cot(\pi x)$, namely, $$\cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi x} \sum_{k=1}^{\infty}\zeta(2k) x^{2k}$$
we have $$S(1)=-\ln(2)+\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k(k+1)(2k+1)}$$
Now since for $\Re(s)>1$ we have the following Mellin transform $$\zeta(s)\Gamma(s)=\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}\,dx$$ we get $$S(1)=-\ln(2)+\int_{0}^{\infty}\frac{1}{e^x-1}\sum_{k=1}^{\infty}\frac{x^{2k-1}}{\Gamma(2k)k(k+1)(2k+1)}=-\ln(2)-2\int_{0}^{\infty}\frac{2+x^2-2\cosh(x)}{x^3(e^x-1)}\,dx$$
We shall now evaluate the integral $$I:=-2\int_{0}^{\infty}\frac{2+x^2-2\cosh(x)}{x^3(e^x-1)}\,dx$$
Write $$\frac{1}{e^x-1}=\sum_{n=1}^{\infty}e^{-nx}$$
so $$I=-2\sum_{n=1}^{\infty}\int_{0}^{\infty} \frac{e^{-n x}}{x^3}\left(2+x^2-2\cosh(x)\right)\,dx$$
Now using the convolution property of the Laplace Transform:
$$\int_{0}^{\infty} f(x)\cdot g(x)\, dx=\int_{0}^{\infty} (\mathcal{L}f)(s)\cdot (\mathcal{L}^{-1}g)(s)\, ds$$ with $f(x) = e^{-nx}\left(2+x^2-2\cosh(x)\right)$ and $g(x)=\frac{-2}{x^3}$ we have $$I=\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{2 s^2}{(n+s-1) (n+s)^3 (n+s+1)}\, ds=\int_{0}^{\infty}\frac{s^2+2 s+2}{s(s+1)}+s^2 \psi ^{(2)}(s)\, ds\stackrel{\text{IBP}}{=}\ln(2\pi)-\frac{3}{2}$$
where $\psi$ is the polygamma function. So $S(1)=\ln(\pi)-\frac{3}{2}$ as required. $\square$
A: Too long for a comment.
What is interesting is that the partial sums
$$S_p=\sum_{n=2}^{p}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$ are given in terms of the gamma function and the derivatives of the zeta function with respect to its first argument.
$$S_p=-\log(2e)+p+\log \big(\Gamma (p)\,\Gamma (p+2)\big)+$$
$$\zeta ^{(1,0)}(-2,p)-2 \zeta ^{(1,0)}(-2,p+1)+\zeta
   ^{(1,0)}(-2,p+2)+2 \zeta ^{(1,0)}(-1,p)-2 \zeta ^{(1,0)}(-1,p+2)$$ Expanded as a series
$$S_p=\log (\pi )-\frac{3}{2}+\frac{1}{2
   p}-\frac{1}{4 p^2}+O\left(\frac{1}{p^3}\right)$$ The same happens with
$$T_p(k)=\sum_{n=2}^{\infty}\left (n^k \ln\left(1-\frac{1}{n^k}\right)+1\right)$$
A: $$\begin{align}\sum^\infty_{n=2}n^2\ln\left(1-\frac{1}{n^2}\right)+1&=\sum^\infty_{n=2}n^2\ln\left(1-\frac{1}{n^2}\right)+\ln(e)=\\&=\sum^\infty_{n=2}\ln\left(1-\frac{1}{n^2}\right)^{n^2}+\ln(e)\;.\end{align}$$ $$\begin{align}\sum^\infty_{n=2}\ln\left(1-\frac{1}{n^2}\right)^{n^2}+\ln(e)&=\sum^\infty_{n=2}\ln\left(\left(1-\frac{1}{n^2}\right)^{n^2}e\right)=\\&=\ln\prod_{n=2}^\infty\left(\frac{n^2-1}{n^2}\right)^{n^2}\!\!e\,.\end{align}$$ Prove that the input of the logarithm is equal to $\pi/e^{3/2}$ and you would be good. Use this paper to get the answer.
