How to calculate $\sum_{n= 0}^{\infty} \left( \frac {n+1} {n+2}+n(n+1)\ln \left(1-\frac 1{(n+1)^2}\right)\right)$

By writing $\log\left(1-\frac{1}{(n+1)^2}\right)=\log n+\log(n+2)-2\log(n+1)$, we get

\begin{align*} &\sum_{n=1}^{N} n(n+1)\log\left(1-\frac{1}{(n+1)^2}\right) \\ &= \sum_{n=1}^{N} n(n+1)\log n - 2 \sum_{n=1}^{N} n(n+1)\log(n+1) + \sum_{n=1}^{N} n(n+1)\log(n+2) \\ &= \Bigg( \sum_{n=1}^{N} n(n+1)\log n - 2 \sum_{n=1}^{N} (n-1)n\log n + \sum_{n=1}^{N} (n-2)(n-1)\log n \Bigg) \\ &\qquad - 2N(N+1)\log(N+1) + N(N+1)\log (N+2) + (N-1)N\log(N+1) \\ &= 2\log(N!) - N(N+3)\log(N+1) + N(N+1)\log(N+2). \end{align*}

Now expanding the last expression for large $N$ and using the Stirling's approximation, we easily check that

\begin{align*} \begin{gathered} 2\log(N!) = (2N+1)\log N - 2N + \log(2\pi) + o(1), \\ - N(N+3)\log(N+1) = -N(N+3)\log N - N - \frac{5}{2} + o(1), \\ N(N+1)\log(N+2) = N(N+1)\log N + 2N + o(1), \end{gathered} \end{align*}

as $N\to\infty$, and so,

\begin{align*} \sum_{n=1}^{N} n(n+1)\log\left(1-\frac{1}{(n+1)^2}\right) = \log N - N + \log(2\pi) - \frac{5}{2} + o(1) \quad \text{as } N \to \infty. \end{align*}

On the other hand,

$$ \sum_{n=0}^{N} \frac{n+1}{n+2} = N+2 - \sum_{n=1}^{N+2} \frac{1}{n} = N+2-(\gamma + \log(N+2) + o(1)) $$

by the definition of the Euler-Mascheroni constant $\gamma$. Combining altogether and letting $N\to\infty$, we get

$$ \sum_{n=0}^{\infty} \left[ \frac{n+1}{n+2} - n(n+1)\log\left(1-\frac{1}{(n+1)^2}\right) \right] = - \frac{1}{2} - \gamma + \log(2\pi). $$

Here, we regard $0\log 0 = 0$. The following numerical computation confirms this:

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Addendum. We may regularize the integral by

$$ I(\alpha, s) := \int_{0}^{\infty} x^{s-1} \left(\frac{1}{x} - \frac{\alpha}{e^{\alpha x} - 1} \right)^2 \, \mathrm{d}x. $$

This integral convergess for $\alpha > 0$ and $s \in (0, 2)$. Also, the substitution $\alpha x \mapsto x$ gives

$$ I(\alpha, s) = \alpha^{2-s} I(1, s).$$

Our goal is to find a closed form of $I(1, s)$ and utilize it to compute the value of $I(1, 1)$. We begin by noting that

$$ \int_{0}^{\infty} \frac{x^{s-1}}{e^{\alpha x} - 1} \, \mathrm{d}x = \frac{\Gamma(s)}{\alpha^s}\zeta(s) $$

for any $\alpha > 0$ and $s > 1$. Differentiating both sides by $\alpha$,

$$ -\int_{0}^{\infty} \frac{x^{s}e^{\alpha x}}{(e^{\alpha x} - 1)^2} \, \mathrm{d}x = -\frac{\Gamma(s+1)}{\alpha^{s+1}}\zeta(s). $$

Replacing $s$ by $s-1$ and subtracting $\int_{0}^{\infty} \frac{x^{s-1}}{e^{\alpha x} - 1} \, \mathrm{d}x$ from both sides,

$$ \int_{0}^{\infty} \frac{x^{s-1}}{(e^{\alpha x} - 1)^2} \, \mathrm{d}x = \frac{\Gamma(s)}{\alpha^{s}}[ \zeta(s-1) - \zeta(s)], $$

which we know to hold for any $\alpha > 0$ and $s > 2$. On the other hand,

\begin{align*} I(1,s) - I(\alpha, s) &= \int_{0}^{\infty} x^{s-1} \bigg[ \left(\frac{1}{x} - \frac{1}{e^{x} - 1} \right)^2 - \left(\frac{1}{x} - \frac{\alpha}{e^{\alpha x} - 1} \right)^2 \bigg] \, \mathrm{d}x \\ &= \int_{0}^{\infty} x^{s-1} \bigg( -\frac{2x^{-1}}{e^{x} - 1} + \frac{1}{(e^{x} - 1)^2} + \frac{2\alpha x^{-1}}{e^{\alpha x} - 1} - \frac{\alpha^2}{(e^{\alpha x} - 1)^2} \bigg) \, \mathrm{d}x. \end{align*}

We notice that this integral converges for all $s > 0$. Moreover, for each $\alpha > 0$ this defines a holomorphic function in $s > 0$. Now by assuming $s > 2$,

\begin{align*} I(1,s) - I(\alpha, s) &= \int_{0}^{\infty} \bigg( -\frac{2x^{s-2}}{e^{x} - 1} + \frac{x^{s-1}}{(e^{x} - 1)^2} + \frac{2\alpha x^{s-2}}{e^{\alpha x} - 1} - \frac{\alpha^2 x^{s-1}}{(e^{\alpha x} - 1)^2} \bigg) \, \mathrm{d}x \\ &= (1 - \alpha^{2-s})\big( \Gamma(s)[ \zeta(s-1) - \zeta(s)] - 2\Gamma(s-1)\zeta(s-1) \big). \end{align*}

Although this holds a priori for $s > 2$, this continues to hold for all of $s > 0$ by the principle of analytic continuation. Comparing this with $ I(1,s) - I(\alpha, s) = (1 - \alpha^{2-s})I(1, s) $, we get

$$ I(1, s) = \Gamma(s)[ \zeta(s-1) - \zeta(s)] - 2\Gamma(s-1)\zeta(s-1) $$

for $s \in (0, 2)$. Using $\zeta(s) = \frac{1}{s-1} + \gamma + \mathcal{O}(s-1)$ as $s \to 1$ and $\zeta(0) = -\frac{1}{2}$, the right-hand side expands as

$$ I(1, s) = \Gamma(s) \bigg( \zeta(s-1) - \gamma - 2 \frac{\zeta(s-1) - \zeta(0)}{s-1} \bigg) + \mathcal{O}(s-1). $$

Letting $s \to 1$, this converges to

$$ I(1, 1) = \zeta(0) - \gamma - 2\zeta'(0) = -\frac{1}{2} - \gamma + \log(2\pi). $$