How to calculate $\sum_{n= 0}^{\infty} \left( \frac {n+1} {n+2}+n(n+1)\ln \left(1-\frac 1{(n+1)^2}\right)\right)$ How to calulate $$\sum_{n= 0}^{\infty} \left( \frac {n+1} {n+2}+n(n+1)\ln \left(1-\frac 1{(n+1)^2}\right)\right)?$$
My goal was to calculate the integral $\int_{0}^{\infty}\left[\frac{1}{x}-\frac{1}{e^x-1}\right]^2 dx$.
Using the three results
1) $\frac{1}{(1-e^{-x})^2}=\sum_{n=0}^{\infty}(n+1)e^{-nx}$
2)$a_n=\int_0^1\int_0^1\frac{1}{(n+u+v)^3}uvdudv=\frac{1}{2(n+2)}+\frac{n}{2}\log\left(1-\frac{1}{(n+1)^2}\right)$
3)$e^x-1-x=e^xx^2\int_0^1ue^{-ux}du$.
We deduce 
$\int_{0}^{\infty}\left[\frac{1}{x}-\frac{1}{e^x-1}\right]^2dx$
=$\int_{0}^{\infty}\frac{x^2}{(1-e^{-x})^2}\left[\int_0^1ue^{-ux}du\int_0^1ve^{-vx}dv\right]dx$
=$\int_0^1\int_0^1\left[\sum_0^{\infty}(n+1)\int_0^{\infty}x^2e^{-x(n+u+v)}dx\right]uvdudv$
= $ 2\int_0^1\int_0^1\left[\sum_0^{\infty}\frac{n+1}{(n+u+v)^3}\right]uvdudv $
= $ 2\sum_0^{\infty}(n+1) a_n  =?$
I'm interested in other ways too, but I'd like to understand how to calculate the sum of this series 
 A: I will use a generalisation of the gamma function, called $Q_m(x)$ in the script here . This script is not perfect, but the formulas I use are correct.
We have on page $13$ the formula $(4.1)$

$$\sum\limits_{n=1}^N n^m \ln\left(1+\frac{x}{n}\right) = p_{m,N}(x) + r_m(x)\ln N - \ln Q_{m,N}(x)$$ 

and additionally  
$\displaystyle Q_m(x):=Q_{m,N}(x)|_{N\to\infty} \enspace, \enspace\enspace Q_m^*(x) := (1+x) Q_m(x) \enspace,$
$\displaystyle r_1(x)=-\frac{x^2}{2} \enspace, \enspace\enspace  r_2(x)=\frac{x^3}{3} \enspace,$
$\displaystyle p_{1,N}(x)=xN \enspace, \enspace\enspace  p_{2,N}(x)=x\frac{N(N+1)}{2}-\frac{x^2}{2}N \enspace$ .
Now we can write 
$\displaystyle \sum\limits_{n=0}^{N-1}\left(x^2\frac{n+1}{n+2}+n(n+1)\ln\left(1-\frac{x^2}{(n+1)^2}\right)\right) = x^2(N+1-H_{N+1}) -$
$\enspace\enspace - (p_{1,N}(-x) + p_{1,N}(x)) + (p_{2,N}(-x) + p_{2,N}(x)) - (r_1(-x) + r_1(x))\ln N $
$\enspace\enspace + (r_2(-x) + r_2(x))\ln N + \ln(Q_{1,N}(-x) Q_{1,N}(x)) - \ln(Q_{2,N}(-x) Q_{2,N}(x))$
$\displaystyle \to ~ x^2(1-\gamma) +  \ln(Q_1(-x)Q_1(x)) - \ln(Q_2(-x)Q_2(x)) \enspace$ for $\enspace N\to\infty$
$\displaystyle = x^2(1-\gamma) +  \ln(Q_1^*(-x) Q_1(x)) - \ln(Q_2^*(-x) Q_2(x))$
Formula $(4.7)$ on page $16$ with $m:=2$ leads to $$\ln(Q_2^*(-1)Q_2(1))=\frac{1}{2}-\ln(2\pi)$$
and formula $(4.16)$ on page $24$ with $m:=1$ gives us $$\ln(Q_1^*(-1)Q_1(1))=-1$$ 
so that with $x:=1$ we get finally: 

$$\sum\limits_{n=0}^\infty\left(\frac{n+1}{n+2}+n(n+1)\ln\left(1-\frac{1}{(n+1)^2}\right)\right) = -\gamma - \frac{1}{2} + \ln(2\pi)$$

