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.

For each $1\leq i\leq n$, $\frac{1}{n^2+i}\leq \frac{1}{n^2}$ and $\frac{1}{n^2+i}\geq \frac{1}{n^2+n}$, and so we may bound the sum from above and below by $$\sum_{i=1}^{n}\frac{i}{n+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{n^{2}}.$$ Since $\sum_{i=1}^{n}i=\frac{n(n+1)}{2},$ this becomes $$\frac{1}{2}=\frac{n(n+1)}{2n(n+1)}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\frac{n(n+1)}{2n^{2}}=\frac{1}{2}\left(1+\frac{1}{n}\right),$$ and so it follows from the squeeze theorem that the limit is $\frac{1}{2}$.
