I have to find the following limit:

> 
> $$\lim\_{n \to \infty} \frac{\frac{\ln 2}{2} + \frac{\ln 3}{3} + \cdots + \frac{\ln n}{n}}{\ln^2n}$$
> 
> 
> 

I tried splitting this limit like so:

$$\lim\limits\_{n \to \infty} \dfrac{\frac{\ln 2}{2} + \frac{\ln 3}{3} + \cdots + \frac{\ln n}{n}}{\ln n} \cdot \dfrac{1}{\ln n}$$

Because $\dfrac{1}{\ln n} \rightarrow 0$ as $n \rightarrow \infty$, I concluded that the limit is $0$. I know that in order to use this I firstly would have to show that

$$\lim\limits\_{n \to \infty} \dfrac{\frac{\ln 2}{2} + \frac{\ln 3}{3} + \cdots +\frac{\ln n}{n}}{\ln n}$$

is bounded, but I didn't know how to do that and kinda hoped for the best. It turns out that my hopes were in vain, since the limit is actually $\dfrac{1}{2}$ and not $0$, like I got. I also tried using Stolz-Cesaro, resulting in:

$$\lim\limits\_{n \to \infty} \dfrac{\frac{\ln (n+1)}{n+1}}{\ln^2 (n + 1) - \ln^2 n} = \lim\limits\_{n \to \infty} \dfrac{\frac{\ln (n+1)}{n + 1}}{(\ln (n+1)-\ln n)(\ln (n + 1) + \ln n)}$$

$$= \lim\limits\_{n \to \infty} \dfrac{\frac{\ln (n + 1)}{n + 1}}{\ln (\frac{n + 1}{n}) \cdot \ln(n(n + 1))}$$

Aaand I got stuck.

So how should I approach this **and** get $\dfrac{1}{2}$ as the final answer?