Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit 
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$ 
My approach : 
If I divide numerator and denominator by $n^2$ I get : 
$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$ 
but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.

Since
$$
1+n^2 \le k+n^2\le n+n^2 \quad \forall k \in \{1,2,\ldots,n\},
$$
it follows that
$$
\frac{n(n+1)}{2(n+n^2)}=\sum_{k=1}^n\frac{k}{n+n^2}\le \sum_{k=1}^n\frac{k}{k+n^2}\le\sum_{k=1}^n\frac{k}{1+n^2}=\frac{n(n+1)}{2(1+n^2)} \quad \forall n \ge 1.
$$
Thus
$$
\lim_n\sum_{k=1}^n\frac{k}{k+n^2}=\frac12.
$$
