Evaluating Limit $$\frac{\ln2}{2}+\cdots+\frac{\ln n}{n}$$

[Vali]

Hi,

$$\lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}.$$
After I applied Stoltz-Cesaro I got $$\lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}=\lim_{n \to \infty}\dfrac{\dfrac{\ln (n+1)}{n+1}}{\ln^2 (n+1)-\ln^2 n}$$
How to continue ? The limit shouldn't be 0 ? because$\lim_{n \to \infty}\frac{ln(n+1)}{n+1}=0$
It's not 0, it's $1/2$ and I don't know why.

 

[Vali]

I did it, thanks!

 

[MountEvariste]

Everything is right except the part where the denominator/numerator limits are taken separately (i.e. the very last part).