Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k}$ I came across this problem that I'm supposed to be able to solve in under 5 minutes (for a competition). 
$$\lim_{n\to \infty} \sum_{k=0}^n \frac {n}{n^2+k}$$
I tried solving this for small sums, $\sum_{k=0}^2 \frac {n}{n^2+k}$, $\sum_{k=0}^3 \frac {n}{n^2+k} $ and made this:
$$\sum_{k=0}^2 \frac {n}{n^2+k} = \frac {n}{n^2+1} + \frac{n}{n^2+2} = \frac {n(n^2+2)+n(n^2+1)}{(n^2+1)(n^2+2)} = \frac {2n^3+...}{n^4+...}$$
And it's limit is $0$, for $k$ up to $3$ is $0$ too, so I assumed that the original limit must be $0$. Wrong, my book says that is $1$, I graphed the function and indeed that is correct. How should I solve it?

$\displaystyle \sum_{k=0}^n \frac {n}{n^2+k} = \frac {n}{n^2+0} + \frac{n}{n^2+1} +.......... +\frac{n}{n^2+n} $
$\displaystyle = \frac {n(n^2+1)(n^2+2)\cdots(n^2+n)+n(n^2)(n^2+2)......}{n^2(n^2+1)(n^2+2)....(n^2+n)} = \frac {n.n^{2n}(n+1)+}{n^2\cdot n^{2n}+}$
$\displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^n \frac {n}{n^2+k}= \lim_{n\rightarrow \infty}\frac {n.n^{2n}(n+1)+....}{n^2\cdot n^{2n}+.....} = 1$
