Show $\lim\limits_{n \to \infty} \frac{1}{n^2} \sum\limits_{k=1}^n \frac{k+1}{\log(k+1)} = 0$

Question

Show that

$$ \lim\limits_{n \to \infty} \frac{1}{n^2} \sum\limits_{k=1}^n \frac{k+1}{\log(k+1)} = 0 $$

I have no clue how to show this. I would presume there is a clever way to use one of the $n$'s inside the sum to cancel appropriately but I can't seem to get the right idea.

Answer 1

By Stolz-Cesaro theorem, $$\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n \frac{k+1}{\log(k+1)}= \lim_{n \to \infty} \frac{\frac{n+2}{\log(n+2)}}{(n+1)^2-n^2} =\lim_{n \to \infty} \frac{1+\frac{2}{n}}{(2+\frac{1}{n})\log(n+2)}=0.$$