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?
 A: Slightly more involved solution: 
$$
\frac{1}{n} \sum_{k=0}^{n} \frac{1}{1+\frac{k}{n^2}} = \frac{1}{n} \sum_{k=0}^{n} e^{- \log (1+\frac{k}{n^2}) } \sim \frac{1}{n} \sum_{k=0}^{n} e^{-\frac{k}{n^2}} = \frac{1}{n} \times \bigg(\frac{1-e^{-\frac{1}{n^2}(n+1)}}{1-e^{-\frac{1}{n^2}}} \bigg) \sim \frac{1}{n} \times \bigg( \frac{\frac{1}{n} + \frac{1}{n^2}}{\frac{1}{n^2}} \bigg) = 1 +\frac{1}{n} \to_n 1
$$
A: Note: $\frac{n}{n^2+n}\leq\frac{n}{n^2+k}\leq\frac{n}{n^2+1}\leq\frac{1}{n}$ for all $k$ between $1$ and $n$. Hence, $\frac{1}{n+1}\leq\frac{n}{n^2+k}\leq\frac{1}{n}$. Note that $\sum_{k=0}^{n}\frac{1}{n}=\frac{n+1}{n}$ and $\sum_{k=0}^{n}\frac{1}{n+1}=\frac{n+1}{n+1}=1$. Therefore: $$1\leq\lim_{n\to\infty}\sum_{k=0}^{n}\frac{n}{n^2+k}\leq\lim_{n\to\infty}\frac{n+1}{n}=1$$
A: $\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} $
$\displaystyle = \frac {n(n^2+1)(n^2+2)\cdots(n^2+n)+n(n^2)(n^2+2)......}{n^2(n^2+1)(n^2+2)....(n^2+n)} = \frac {n.n^{2n}(n+1)+}{n^2\cdot n^{2n}+}$
$\displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^n \frac {n}{n^2+k}= \lim_{n\rightarrow \infty}\frac {n.n^{2n}(n+1)+....}{n^2\cdot n^{2n}+.....} = 1$
A: For every value of n, there are n+1 terms, the largest of which is ${n\over n^2}={1\over n}$.  Therefore, the sum is always $\le {n+1\over n}$ which converges to 1.  
Similarly, the smallest term in each sum is ${n\over n^2+n}={1\over n+1}$, so the sum is always $\ge {n+1\over n+1}=1$ which also converges to 1.
Since the desired sum is both $\ge$ and $\le$ sums which converge to 1, it also converges to 1.
