Find $\lim_{n \to \infty}\frac1{\ln^2n}\left( \frac{\ln 2}{2} + \frac{\ln 3}{3} +\cdots + \frac{\ln n}{n}\right)$ 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?
 A: Define
$$
S_n=\frac{\ln 2}{2} + \frac{\ln 3}{3} + ... + \frac{\ln n}{n}
$$
The function $\ln(x)/x$ is decreasing for $x\geq 3$, so
$$
\frac{\ln 2}{2} + \int_3^n\frac{\ln(x)}{x}\,dx  \leq S_n\leq \frac{\ln 2}{2} + \frac{\ln 3}{3} + \int_3^{n}\frac{\ln(x)}{x}\,dx,
$$
$$
\frac{\ln(2)}{2} + \frac{\ln(n)^2}{2}-\frac{\ln(3)^2}{2}\leq S_n\leq\frac{\ln(2)}{2}+\frac{\ln(3)}{3} + \frac{\ln(n)^2}{2}-\frac{\ln(3)^2}{2}.
$$
Now divide through by $\ln(n)^2$ and apply the squeeze theorem to conclude
$$
\lim_{n\to\infty}\frac{S_n}{\ln(n)^2}=\frac{1}{2}.
$$
A: Your approach is correct. At the end you need to rewrite your expression as $$\dfrac{\log n+\log\left(1+\dfrac{1}{n}\right)}{\dfrac{n+1}{n}\cdot n\log\left(1+\dfrac{1}{n}\right)\cdot \left(2\log n+\log\left(1+\dfrac{1}{n}\right)\right)}$$ Dividing the numerator and denominator by $\log n$ you can easily see that the numerator tends to $1$ and denominator tends to $2$ and you are done.

We have made use of the fact that $\log n\to\infty$ and $n\log(1+(1/n))\to 1$ as $n\to\infty $. 
