<s>[INST] 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?
 [/INST] 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: $$1\leq\lim_{n\to\infty}\sum_{k=0}^{n}\frac{n}{n^2+k}\leq\lim_{n\to\infty}\frac{n+1}{n}=1$$
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