To show that the limit of the sequence $\sum\limits_{k=1}^n \frac{n}{n^2+k^2}$ is $\frac{\pi}{4}$

Question

Show that $$\lim_{n \to \infty} \sum\limits_{k=1}^n \frac{n}{n^2+k^2} = \frac{\pi}{4}.$$

I am familiar with Taylor series and Fourier series of the standard functions. I tried to compare with those and see if there is a relation, I don't seen to find any. How do I tackle this?

Answer 1

We have $$\sum_{k=1}^{n}\frac{n}{n^2+k^2}=\sum_{k=1}^{n}\frac{1}{1+(k/n)^2}\left(\frac{1}{n}\right)\to \int_0^1\frac{1}{1+x^2}\,dx=\frac{\pi}{4}$$

Answer 2

Notice, $$\lim_{n\to \infty}\sum_{k=1}^{\infty}\frac{n}{n^2+k^2}$$ $$=\lim_{n\to \infty}\sum_{k=1}^{\infty}\frac{n}{n^2\left(1+\left(\frac{k}{n}\right)^2\right)}$$ $$=\lim_{n\to \infty}\sum_{k=1}^{\infty}\frac{\left(\frac{1}{n}\right)}{1+\left(\frac{k}{n}\right)^2}$$ let $\frac{k}{n}=u$, $\frac{1}{n}=du$ as $n\to \infty$

upper limit of $u$ at $r=n$, $\lim_{n\to \infty}\frac{r}{n}=\lim_{n\to \infty}\frac{n}{n}=1$

lower limit of $u$ at $r=1$, $\lim_{n\to \infty}\frac{r}{n}=\lim_{n\to \infty}\frac{1}{n}=0$ using integration with proper limits, $$\lim_{n\to \infty}\sum_{k=1}^{\infty}\frac{\left(\frac{1}{n}\right)}{1+\left(\frac{k}{n}\right)^2}=\int_{0}^{1}\frac{du}{1+u^2}$$ $$=[\tan^{-1}(u)]_{0}^{1}=\tan^{-1}(1)-\tan^{-1}(0)=\color{red}{\frac{\pi}{4}}$$