User:Fredrik/Pretty formulas

Gamma function series

 * $$\frac{\sqrt\pi}{2} = \lim_{n\to\infty} \frac{\sqrt n}{e^n} \sum_{k=0}^\infty \frac{(2n)^k}{(2k+1)!!}$$


 * $$\frac{\Gamma(1/p)}{p} = \lim_{n\to\infty} \frac{\sqrt[p] n}{e^n} \sum_{k=0}^\infty \frac{(pn)^k}{(pk+1)!_{(p)}}\,$$


 * $$\Gamma(z+1) = \sum_{n=1}^\infty \left[ \, \sum_{m=1}^{2^n-1} \, (-1)^{m+1} \frac{(n \ln 2 - \ln m)^z}{2^n} \right]$$

Trigonometric series
Divergent series that can be evaluated with Mathematica (and some trickery):


 * $$\sum_{n=1}^\infty \frac{i^n}{n} = -\frac{\log 2}{2}+\frac{\pi}{4} \, i$$


 * $$\sum_{n=1}^\infty \frac{\sin n}{n} = \frac{\pi-1}{2}$$


 * $$\sum_{n=1}^\infty \frac{\cos n}{n} = \frac{-\log(2-2\cos 1)}{2}$$

The divergent series


 * $$A = \sum_{n=1}^\infty \sin n$$

can be evaluated as a geometric series:


 * $$A = \mathrm{Im} \left[ \sum_{n=1}^\infty e^{n i} \right] = \mathrm{Im} \left[ -\frac{e^i}{e^i - 1} \right] = \left(2 \tan \frac{1}{2}\right)^{-1} = 0.91524386...$$

This number can also be written


 * $$A = 1 + \sum_{n=1}^\infty (-1)^n \frac{B_{2n}}{(2n)!} = 1 - 2 \sum_{n=1}^\infty \frac{\zeta(2n)}{(2 \pi)^{2n}}.$$

Likewise,


 * $$B = \sum_{n=1}^\infty \cos n = \mathrm{Re} \left[ -\frac{e^i}{e^i - 1} \right] = -\frac{1}{2}$$

Interpretation: A and B are "averages", in the following sense: If


 * $$a_m = \sum_{n=1}^m \sin n$$ and $$b_m = \sum_{n=1}^m \cos n,$$

then


 * $$A = \frac{1}{2}\left(\max(a_1, a_2, \ldots) + \min(a_1, a_2, \ldots)\right)$$

and
 * $$B = \frac{1}{2}\left(\max(b_1, b_2, \ldots) + \min(b_1, b_2, \ldots)\right).$$