How this series is bounded

Question

I have

$\sum_{n=2}^\infty \frac{\log n}{n(n-1)}= O(1)$

Do you have any idea why it is bounded?

Answer 1

It it bounded because it is a convergent series. Indeed, remember $\log n=o\bigl(n^{1/2}\bigr)$ and $n^2-n\sim_\infty n^2$, so $$\frac{\log n}{n(n-1)}\sim_\infty \frac{\log n}{n^2}=o\biggl(\frac{n^{1/2}}{n^2}\biggr)=o\biggl(\frac1{n^{3/2}}\biggr),$$ and the series $\displaystyle \sum_{n\ge 1}\frac1{n^{3/2}}$ converges.

Answer 2

For any $n\geq 2$ we have $$ \frac{\log n}{n}-\frac{\log(n+1)}{n+1} = \frac{\log(n+1)}{n(n+1)}-\frac{\log\left(1+\frac{1}{n}\right)}{n}\tag{1} $$ and by exploiting $\log\left(1+\frac{1}{n}\right)\leq \frac{1}{n}$ and rearranging $$ \frac{\log n}{n(n+1)}\leq \left(\frac{\log n}{n}-\frac{\log(n+1)}{n+1}\right) + \frac{1}{n^2}\tag{2} $$ such that, by creative telescoping, $$ \sum_{n\geq 2}\frac{\log n}{n(n-1)} = \frac{\log 2}{2}+\sum_{n\geq 2}\frac{\log(n+1)}{n(n+1)}\leq \log(2)+\zeta(2)-1.\tag{3} $$ We may exploit Frullani's integral to derive an exact integral representation of the LHS: $$ \sum_{n\geq 2}\frac{\log n}{n(n-1)} = -\int_{0}^{+\infty}(1-e^{-x})\log(1-e^{-x})\frac{dx}{x}=\int_{0}^{1}\frac{(1-u)\log(1-u)}{u\log u}\,du \tag{4}$$ and numerical methods to get LHS $\approx 1.25774688694437$.