Proving that this series has a finite sum Consider the following series
$$\sum_{n=1}^{\infty}\dfrac{\log n}{n(n-1)}$$
I have tried to use the ratio test, but then I would get
$$\dfrac{(n-1)\log(n+1)}{(n+1)\log n}$$
And taking the limit as $n \to \infty$ would yield 1 so I don't think it would help.
 A: Note that $\log(n)\le \sqrt{n}$.  Hence, we have
$$\left|\frac{\log(n)}{n(n-1)}\right|\le \frac{1}{n^{1/2}(n-1)}$$
Can you finish now?
A: Hint. The Ratio Test is inconclusive here. Note that if $a\in(1,2)$ then
$$\lim_{n\to \infty}\frac{\dfrac{\log n}{n(n-1)}}{\dfrac{1}{n^{a}}}=\lim_{n\to \infty}\dfrac{n^a\log n}{n(n-1)}=0$$
which implies that there is an $N>0$ such that for all $n\geq N$,
$$0\leq \dfrac{\log n}{n(n-1)}\leq \dfrac{1}{n^{a}}.$$
What may we conclude?
A: I guess the sum starts at $n=2$, otherwise one should define the meaning of $\frac{\log(n)}{n(n-1)}$ at $n=1$.
By Frullani's theorem we have $\log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx$, and since $\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}$, by the dominated convergence theorem it follows that
$$ \sum_{n\geq 2}\frac{\log(n)}{n(n-1)} = -\int_{0}^{+\infty}(1-e^{-x})\log(1-e^{-x})\frac{dx}{x}\stackrel{x\mapsto -\log u}{=}\int_{0}^{1}\frac{(1-u)\log(1-u)}{u\log u}\,du.$$
By expanding $\frac{1}{n(n-1)}$ as a geometric series we also have:
$$ \sum_{n\geq 2}\frac{\log(n)}{n(n-1)} = -\left(\zeta'(2)+\zeta'(3)+\zeta'(4)+\zeta'(5)+\ldots\right) $$
The RHS is clearly convergent since $0\leq -\zeta'(s) \leq \frac{1}{(s-1)^2}$ for any $s>1$. In particular the LHS is non negative but bounded by $\zeta(2)$. A simpler proof of convergence comes from noticing that, for any $n\geq 2$,
$$ \log(n)=\int_{1}^{n}\frac{dx}{x}\stackrel{\text{CS}}{\leq}\sqrt{(n-1)\int_{1}^{n}\frac{dx}{x^2}} = \sqrt{n}-\frac{1}{\sqrt{n}}$$
hence:
$$ \sum_{n\geq 2}\frac{\log(n)}{n(n-1)} \leq -1+\zeta\left(\tfrac{3}{2}\right). $$
As a further alternative, it is enough to notice that both $\frac{u-1}{\log u}$ and $\frac{-\log(1-u)}{u}$ are non-negative functions in $L^2(0,1)$.
