Prove that $\frac{1}{n} \sum_{k=2}^n \frac{1}{\log k}$ converges to $0$ 
Prove that $\frac{1}{n} \sum_{k=2}^n \frac{1}{\log k}$ converges to $0.$

Okay, seriously, it's like this question is mocking me.  I know it converges to $0$.  I can feel it in my blood.  I even proved it was Cauchy, but then realized that didn't tell me what the limit was.  I've been working on this for an hour, so can one of you math geniuses help me?
Thanks!

In general, if $a_n\to 0$, then $\frac1n \sum_{k=0}^n a_k \to 0$ too.
(For any $\varepsilon>0$, find an $N$ such that $|a_n|<\varepsilon/2$ for all $n>N$, and then take enough terms beyond $N$ that they dominate whatever the terms before $N$ contribute to the average).
Even more generally (and as an easy consequence), whenever $a_n$ converges, $\frac1n \sum_{k=0}^n a_k$ converges to the same limit.
