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

Using real analysis techniques.

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Riemann integral sound familiar? –  Ron Gordon Oct 31 '13 at 0:36

3 Answers 3

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

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

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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 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.

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