$ \lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$ Evaluate:
$$ L=\lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$$
My approach:
Each term can be written as
$$ \frac k{n^2+k}=\frac {n^2+k-n^2}{n^2+k}=1-\frac {n^2}{n^2+k}$$
$$ \therefore \lim_{n \to \infty}\frac k{n^2+k}=\lim_{n \to \infty}\left(1-\frac {n^2}{n^2+k}\right)=0$$ 
hence,
$$ L=0$$
Problem:
The correct answer is 1/2, please indicate the flaw in my approach or post a new solution.
Thank You

HINT:
Using $\sum_{k=1}^n k=\frac{n(n+1)}{2}$ along with the estimates $n^2+1\le n^2+k\le n^2+n$ reveals
$$\frac{n(n+1)}{2(n^2+n)}\le\sum_{k=1}^n\frac{k}{n^2+k}\le \frac{n(n+1)}{2(n^2+1)}$$
