Convergence of logarithmic sum Numerical experimentations strongly suggest that the series $$\sum_{n=2}^{\infty}\frac{1}{n}\log\left(\frac{n}{n-1}\right)$$ converges and the limit is $\approx0\mathrm.7885$. Could someone give me a hint on a formal proof of convergence?
 A: By summation by parts,
$$\sum_{n\geq 1}\frac{\log(n+1)-\log(n)}{n+1} = \sum_{n\geq 2}\frac{\log n}{n(n+1)}=2\sum_{n\geq 2}\frac{\log\left(n!\right)}{n(n+1)(n+2)}$$
and the series in the middle term is obviously convergent. It can be written as:
$$ \left.\frac{d}{ds}\int_{0}^{1}\left(-x+\text{Li}_{1-s}(x)\right)\,dx\;\right|_{s=0^+}$$
or as:
$$ -\zeta'(2)+\zeta'(3)-\zeta'(4)+\zeta'(5)+\ldots\approx \color{red}{0.78853}. $$
$\log n<\sqrt{\log(n)\log(n+1)}$ and the Cauchy-Schwarz inequality give:
$$ \sum_{n\geq 2}\frac{\log n}{n(n+1)}<-\zeta'(2) $$
but that is not much stronger than the trivial:
$$ \sum_{n\geq 1}\frac{\log\left(1+\frac{1}{n}\right)}{n+1}<\sum_{n\geq 1}\frac{1}{n(n+1)}=1.$$
A: hint: $a_n < \dfrac{1}{n(n-1)}$ 
A: Note that $$\frac{1}{n}\log\left(\frac{n}{n-1}\right) = \frac{1}{n}\log\left(1+\frac{1}{n-1}\right) \sim \frac{1}{n^2}$$
That's enough to intuitively confirm convergence. 
The bound $\frac{1}{n}\log\left(1+\frac{1}{n-1}\right)\leq \frac{1}{(n-1)^2}$ gives a formal proof.
A: The straight forward Limit Comparison Test should also work. Using LeGrandDODUM's comment, use the convergent Series $Bn=\frac{1}{n^2}$ and verify through LCT that you are essentially taking a limit $n ln(\frac{n}{n-1})$ where n goes to infinity. Since this is essentially an "infinity times zero" situation, but the $n$ term in the denominator as $\frac{1}{n}$ and proceed with L'Hospital's Rule to find (after some elementary algebra) the limit going to 1. This confirms the given series to be convergent. 
