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!

Stolz Cezaro:
$$\lim \frac{1}{n} \sum_{k=2}^n \frac{1}{log k} = \lim \frac{1}{ \log (n+1)}$$
Edit
Here is a direct proof:
$$0 \leq  \frac{\sum_{k=2}^n \frac{1}{log k}}{n}  = \frac{\sum_{k=2}^{\sqrt{n}} \frac{1}{log k}}{n}
+ \lim \frac{\sum_{k=\sqrt{n}}^n \frac{1}{log k}}{n} \leq \frac{\sqrt{n} \frac{1}{log 2}}{n}
+ \frac{(n-\sqrt{n})\frac{1}{\log \sqrt{n}}}{n} $$
Now each of the last two sequences converge to $0$, so squeeze it.
