User:LindySoul/sandbox

$$(a+b)^n = \sum_{i=0}^n \left(\stackrel{n}{i}\right) a^{n-i} b^i$$, if and only if n is a positive integer.

Headline text
To find the area of any regular polygon $$A = \frac{l^2 n}{4 \tan{\frac{\pi^\mathrm{R}}{n}}}$$

Apparently
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \mathrm{...} = \sum_{i=1}^\infty \frac{1}{i^2} = \frac{\pi^2}{6}$$

But if $$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \mathrm{...} =\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \mathrm{...} + \frac{1}{(\infty -2)^2} + \frac{1}{(\infty -1)^2} + \frac{1}{\infty^2} $$

And $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} + \frac{1}{d} = \frac{bcd +acd + abd +abc}{abcd}$$

Therefore it seems that through the same pattern: $$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \mathrm{...} = \frac{\infty!/1 + \infty!/2 + \infty!/3 + \mathrm{...}}{1^22^23^2\mathrm{...}}$$