Is there a closed form solution to $\sum_{n=1}^{\infty}\frac{\ln\left(\frac{n+1}{n}\right)}{n}$

Question

While coming up with an idea for another way to milk the integral in my previous question, I got stuck at this summation: $$\sum_{n=1}^{\infty}\frac{\ln\left(\frac{n+1}{n}\right)}{n}$$

I do not know how to approach this summation and Wolfram Alpha does not give a closed form. It is approximately equal to $1.25771469823$. If it is any help, it is the solution to $\int_{0}^{1}\frac{1}{x\operatorname{floor}\left(\frac{1}{x}\right)}dx$.

Answer 1

Expanding the $\ln\left(1+\frac 1n\right)$ terms allows to rewrite your series as $\;\displaystyle \tag{1}S=\sum_{n=1}^\infty (-1)^{n+1}\frac {\zeta(n+1)}n$
(or $\;\displaystyle S=\log 2+\sum_{n=1}^\infty (-1)^{n+1}\frac {\zeta(n+1)-1}n$ for faster convergence)

We may too use the generating function for $\zeta$ ($\psi$ is the digamma function) $$\tag{2}\psi(1+x)=-\gamma-\sum_{n=1}^\infty \zeta(n+1)\;(-x)^n$$ to rewrite your sum as an integral : $$\tag{3} S=\int_0^1 \frac{\psi(1+x)+\gamma}{x} dx$$

(I don't know a closed form but see this thread or the paper Series Involving Euler’s Eta (or Dirichlet Eta) Function )

Answer 2

Since absolutely convergent by comparison to $\sum\frac 1{n^2}$

You can remove the telescopic part i.e. $\sum\left(\frac{\ln(n+1)}{(n+1)}-\frac{\ln(n)}{n}\right)$

To conserve $\sum\frac{\ln(n)}{n(n+1)}$ and apparently you can express this in term of $\zeta'$ see this but still an infinite sum...

See also this page on Alladi-Grinstead Constant