# Show that $\lim_{n\to \infty} \sum_{k=1}^n \frac n{n^2+k^2}=\frac \pi 4$ [duplicate]

My Work: $$\lim_{n\to \infty} \sum_{k=1}^n \frac n{n^2+k^2}$$ $$\lim_{n\to \infty} n\sum_{k=1}^n \frac 1{n^2+k^2}$$ $$\lim_{n\to \infty} n(\frac1{n^2+1^2}+\frac1{n^2+2^2}+\frac1{n^2+3^2}+...+\frac1{n^2+(n-1)^2}+\frac1{2n^2})$$

I'm not really sure where to proceed from here, or if this even is the correct direction that I should take this problem. My math Analysis class has covered up to Riemannian integrals, but I'm not sure that they will be entirely helpful in this instance.

## marked as duplicate by kingW3, Servaes, tired, user147263, user99914 Nov 24 '15 at 12:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• hint: pull out the $n^2$ in the denominator and use the definition of the riemann integral.\ – tired Nov 23 '15 at 16:22

## 2 Answers

$$\lim_{n\to\infty}\sum_{k\to1}^{n}\frac{n}{n^2+k^2}$$ $$=\lim_{n\to\infty}\sum_{k\to1}^{n}\frac{1}{1+(\frac{k}{n})^2}\frac{1}{n}$$ $$=\int_0^1 \frac{1}{1+x^2}dx$$ $$=\tan^{-1}(1)=\frac{\pi}{4}$$

$$\sum\limits_{k=1}^n \frac{n}{n^2+k^2} = \frac{1}{n} \sum\limits_{k=1}^n \frac{1}{1+\left(\frac{k}{n}\right)^2}$$

So $$\lim\limits_{n\to \infty} \sum\limits_{k=1}^n \frac{n}{n^2+k^2} = \int\limits_0^1 \frac{1}{1+x^2} dx = \arctan(1) - \arctan(0) = \frac{\pi}{4}$$