Sum of $\sum_{n=1}^\infty\frac{\ln(n) -\ln(n+1)}{n+1}$

Question

The sum series exercise started as: $$\sum_{n=1}^\infty\frac{\ln\left(\frac{n^n}{\left(n+1\right)^n}\right)}{n(n+1)} = \sum_{n=1}^\infty \frac{n\ln\frac{n}{n+1}}{n(n+1)} = \sum_{n=1}^\infty \frac{\ln(n) - \ln(n+1)}{n+1}$$ Looking into calculating it. A quick hint will help me out!

Answer 1

As $n \to \infty$, the general term of the series satisfies $$ \frac{\ln\left(\frac{n^n}{(n+1)^n}\right)}{n(n+1)}=-\frac{1}{n+1}\ln{\small{\left(1+\frac1n\right)}}\sim -\frac1{n^2} $$ giving the convergence of the series.

From $$ \sum_{n=1}^\infty\frac{\ln\left(\frac{n^n}{(n+1)^n}\right)}{n(n+1)}=\sum_{n=1}^\infty\frac{\ln n-\ln(n+1)}{n+1} $$ one may use Theorem 2 (16) to get a closed form in terms of poly-Stieltjes constants:

$$ \sum_{n=1}^\infty\frac{\ln\left(\frac{n^n}{(n+1)^n}\right)}{n(n+1)}=\gamma_1(0,1)-\gamma_1 \tag1 $$

where $$ \gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right) $$ and $\gamma_1=\gamma_1(1,1)$ is an ordinary Stieltjes constant.

By using Mathematica, one gets

$$ \sum_{n=1}^\infty\frac{\ln\left(\frac{n^n}{(n+1)^n}\right)}{n(n+1)}=-0.7885205660\cdots. \tag2 $$