$$ \lim_{n\to\infty} \frac{1}{n} \sum\limits_{i=1}^{n} \frac{1}{1 + ( \frac{i}{n})^2} $$
I know what this is as an integral, and it's supposed to evaluate to $\frac{\pi}{4}$. I'm not sure how to evaluate this as a limit though, and I want to know because I think it will broaden my understanding a bit.
Note that it is a Riemann sum for $\int_0^1\frac{1}{1+x^2}dx$:
Since $f:[0,1]\to\mathbb{R}$ given by $\frac{1}{1+x^2}$ is continuous, then it is integrable. Then, for any tagged partition $P=\{x_0=0,x_1,...,x_n=1\}$ with tags $x_{i_1}\le t_i\le x_i$, we have $$\int_0^1\frac{1}{1+x^2}dx=\lim_{\|P\|\to0}\sum_{i=1}^nf(t_i)(x_i-x_{i-1}).$$
In particular, if we take $P=\{0,1/n,2/n,...,n/n\}$ and $t_i=x_i$, $i\ge1$, we get $$\int_0^1\frac{1}{1+x^2}dx=\lim_{n\to\infty}\sum_{i=1}^nf(\frac{i}{n})\frac{1}{n},$$ which is your sum.
Thus, the limit is $\frac{\pi}{4}$.
