The sum of logarithmic series

Question

I will be very grateful for help and suggestions how to calculate the sum $$\sum\limits_{n=2}^{\infty}\frac{\log(n)}{n(n-1)}$$

Answer 1

Playing around.

$\begin{array}\\ \sum\limits_{n=2}^{m}\frac{\log(n)}{n(n-1)} &=\sum\limits_{n=2}^{m}\log(n)(\frac1{n-1}-\frac1{n})\\ &=\sum\limits_{n=2}^{m}\log(n)\frac1{n-1}-\sum\limits_{n=2}^{m}\log(n)\frac1{n}\\ &=\sum\limits_{n=1}^{m-1}\log(n+1)\frac1{n}-\sum\limits_{n=2}^{m}\log(n)\frac1{n}\\ &=\sum\limits_{n=2}^{m-1}\frac{\log(n+1)-\log(n)}{n}+\log(2)-\frac{\log(m)}{m}\\ &=\sum\limits_{n=2}^{m-1}\frac{\log(1+1/n)}{n}+\log(2)-\frac{\log(m)}{m}\\ &=\sum\limits_{n=2}^{m-1}\frac1{n}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{kn^k}+\log(2)-\frac{\log(m)}{m}\\ &=\sum_{k=1}^{\infty}\sum\limits_{n=2}^{m-1}\frac{(-1)^{k-1}}{kn^{k+1}}+\log(2)-\frac{\log(m)}{m}\\ &=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum\limits_{n=2}^{m-1}\frac{1}{n^{k+1}}+\log(2)-\frac{\log(m)}{m}\\ &=\sum_{k=2}^{\infty}\frac{(-1)^{k}}{k-1}\sum\limits_{n=2}^{m-1}\frac{1}{n^{k}}+\log(2)-\frac{\log(m)}{m}\\ &\to \sum_{k=2}^{\infty}\frac{(-1)^{k}}{k-1}(\zeta{k}-1)+\log(2)\\ &= \sum_{k=2}^{\infty}\frac{(-1)^{k}}{k-1}\zeta(k)-\sum_{k=2}^{\infty}\frac{(-1)^{k}}{k-1}+\log(2)\\ &= \sum_{k=2}^{\infty}\frac{(-1)^{k}}{k-1}\zeta(k)\\ \end{array} $

At this point, I don't know what to do so I'll leave it here.

Answer 2

Both answers by Mark Viola and marty cohen do not leave much hope for a closed form formula.

You can get a lower bound of the value writing $$\sum\limits_{n=2}^{\infty}\frac{\log(n)}{n(n-1)} > \int_{2}^{\infty}\frac{\log(n)}{n(n-1)}\,dn=\frac{1}{12} \left(\pi ^2+6 \log ^2(2)\right)\approx 1.06269 $$

Computing the summation with illimited precision, the result is $\approx 1.257746886944369$ which is not recognized by inverse symbolic calculators. The closest numer they found is $$\frac 1 {10}\sqrt{23 \left(2^{3/4}+3 \sqrt{3}\right)}\approx 1.257746957$$ which, I bet, is absolutely of no use.

Answer 3

A numerical value is provided in https://oeis.org/A131688 , which also gives the formulas shown above.

Answer 4

By Frullani's integral $$\sum_{n\geq 1}\frac{\log(n+1)}{n(n+1)}=-\int_{0}^{+\infty}\frac{(1-e^{-x})\log(1-e^{-x})}{x}\,dx=\int_{0}^{1}\frac{(1-u)\log(1-u)}{u\log u}\,du $$ hence in terms of Gregory coefficients $$ \sum_{n\geq 1}\frac{\log(n+1)}{n(n+1)}=\zeta(2)+\sum_{n\geq 1}\underbrace{|G_n|\left(H_n^{(2)}-\zeta(2)\right)}_{\sim\frac{C}{n^2\log^2 n}}=\sum_{n\geq 1}|G_n|H_n^{(2)}. $$ and by the integral representation for Gregory coefficients the RHS equals $$ \int_{0}^{+\infty}\frac{1+\frac{1}{x}}{\pi^2+\log^2 x}\,\text{Li}_2\left(\frac{1}{x+1}\right)\,dx=\int_{0}^{1}\frac{\text{Li}_2(u)\,du}{u^2(1-u)(\pi^2+\log^2\frac{1-u}{u})} $$

but the integral representation (got by integration by parts) $$\zeta(2)-\int_{0}^{1}\underbrace{\frac{1-u+u\log u}{u\log^2 u}\,\text{Li}_2(u)}_{g(u)}\,du \tag{R}$$ is best suited for numerical purposes, since $g(u)$ is smooth and bounded on $(0,1)$. For instance, by approximating $g(u)$ with $\frac{\pi^2}{12}u$ we get that the value of the original series is not very far from $\frac{\pi^2}{8}$.
By approximating $\log(n+1)$ with $$ \frac{1}{2}\left(H_n+H_{n+1}\right)-\gamma+\frac{1}{12n(n+1)} $$ we get that the value of the original series is also close to $\frac{\pi^2}{9}+\frac{3}{4}-\gamma$.
Improving this approach through $$ \log(n+1)\approx \tfrac{5}{12}H_n+\tfrac{2}{3}H_{n+1}-\tfrac{1}{12}H_{n+2}-\gamma+\frac{n}{12(n+1)^4} $$ we get $$ S=\sum_{n\geq 1}\frac{\log(n+1)}{n(n+1)}\approx \frac{1}{144} \left(153-144 \gamma +10 \pi^2+12\,\zeta(5)\right)\approx \color{green}{1.25}708 $$

Indeed, Romberg's method applied to ($\text{R}$) produces $S\approx 1.257746887$.