Series calculation using a Riemann Sum and FTC: $\lim_{n \rightarrow \infty}\sum_{k=1}^n \frac{n}{n^2+k^2}$ Any tips on how to do this problem using Reimann Sums and the Fundamental Theorem of Calculus?
$$\lim_{n \rightarrow \infty}\sum_{k=1}^n \frac{n}{n^2+k^2}$$
I got to this stage:
$$\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^n \frac{n^2}{n^2+k^2}$$
$$\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k^2}{n^2}}$$
Is the $f(x)$ actually $\frac{1}{1+x^2}$?
 A: So far so good! All you need to do now is divide through by $n^2$ as follows:
$$ \lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^n \frac{n^2}{n^2+k^2}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^n\frac{1}{1+(\frac{k}{n})^2}.$$
If we think about it, we realize that this is the Riemann Sum corresponding to 
$$ \int_0^1\frac{1}{1+x^2}dx=\arctan x\bigg|_0^1=\arctan 1=\frac{\pi}{4}.$$
To see that this is in fact the Riemann Sum, note that the $\frac{1}{n}$ denotes the "bin" size with which we partition our interval of integration. Here, we note that as $0\le k\le n$, $0\le \frac{k}{n}\le1 $. So, our sample points are ranging from $0$ to $1$. As $n\to \infty$, the bin size (or mesh of the partition) tends to $0$, while the number of sample points tends to infinity. This is precisely the setup for the Riemann Integral.
A: You almost have it. 
\begin{equation}\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k^2}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^n\frac{1}{1+(\frac{k}{n})^2}=\int_0^1\frac{1}{1+x^2}dx=\arctan(x)|_0^1=\frac{\pi}{4}.
\end{equation}
