Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

Using real analysis techniques.

share|cite|improve this question
Riemann integral sound familiar? – Ron Gordon Oct 31 '13 at 0:36

Hint: It's a Riemann sum. $$ \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + (k/n)^2} $$

share|cite|improve this answer

enter image description here

Riemann summation factor n^2 from n^2+k^2 and simplify then go on

share|cite|improve this answer
You can't write limit of something if you don't know if this limit exist...and it's a waste of time :) – Shadock Dec 16 '14 at 17:48

No rigourous, but leads to the solution:

$$ \frac{1}{n}\sum_{k=1}^n\frac1{1+\left(\frac kn\right)^2}\approx\frac{1}{n}\sum_{k=1}^n 1 - (k/n)^2 + (k/n)^4 - (k/n)^6\dots \approx\frac1n(n-\frac{n^3}{3n^2}+\frac{n^5}{5n^4}-\frac{n^7}{7n^6}\dots) ,$$ by keeping the first terms of the Faulhaber sums.

$$1-\frac13+\frac15-\frac17...=\frac\pi4$$ is Gregory's series.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.