Series calculation using a Riemann Sum and FTC: $\lim_{n \rightarrow \infty}\sum_{k=1}^n \frac{n}{n^2+k^2}$

Question

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 $\frac{1}{1+x^2}$?

Answer 1

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.

Answer 2

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}