<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] Slightly more involved solution: 
$$
\frac{1}{n} \sum_{k=0}^{n} \frac{1}{1+\frac{k}{n^2}} = \frac{1}{n} \sum_{k=0}^{n} e^{- \log (1+\frac{k}{n^2}) } \sim \frac{1}{n} \sum_{k=0}^{n} e^{-\frac{k}{n^2}} = \frac{1}{n} \times \bigg(\frac{1-e^{-\frac{1}{n^2}(n+1)}}{1-e^{-\frac{1}{n^2}}} \bigg) \sim \frac{1}{n} \times \bigg( \frac{\frac{1}{n} + \frac{1}{n^2}}{\frac{1}{n^2}} \bigg) = 1 +\frac{1}{n} \to_n 1
$$
</s>