The beauty of Infinity

Infinity and its infinite beauty.

Today we're going to take a brief look at the beauty of infinity, through:

to represent some of the most well known and beautiful mathematical constants, like π, e and φ.

# Definitions

Before diving into it, let's begin with some simple definitions:

1) Infinite sum:

$\sum_{n=1}^(\infty)\frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \...$

2) Infinite product:

$\prod_{n=1}^(\infty)\frac{1}{n} = \frac{1}{1} * \frac{1}{2} * \frac{1}{3} * \...$

3) Continued fractions:

$\underset{n=1}{\overset{\infty}{\mathbf{K}}}\frac{1}{n} = \frac{1}{1 + \frac{1}{2 + \frac{1}{3 + \...}}}$

4) Nested radicals:

$\underset{n=1}{\overset{\infty}{\mathbf{R}}}(n + R)^(1/2) = \sqrt{1 + \sqrt{2 + \sqrt{3 + \...}}}$

5) Limit:

$\lim_{n to \infty}(1 + 1/n)^n = (1 + 1/n)^n$ , where

$n$ approaches

$\infty$ .

# Infinite sums

$\sum_{n=0}^(\infty)\frac{1}{n!} = \e$

$\sum_{n=0}^(\infty)\frac{x^n}{n!} = \e^x$

$\sum_{n=0}^(\infty)\frac{n^x}{n!} = \e * Bell_x$

$\sum_{n=0}^(\infty)\frac{x * (-1)^n}{n!} = \frac{x}{\e}$

$\sum_{n=0}^(\infty)\frac{n!}{(2n+1)!!} = \frac{\pi}{2}$

$\sum_{n=0}^(\infty)\frac{n!*(2n)!!}{(2n+1)!} = \frac{\pi}{2}$

$\sum_{n=0}^(\infty)\frac{n!}{(2n+2)!!} = \log(2)$

$\sum_{n=0}^(\infty)\frac{1}{(2n+2)*2^n} = \log(2)$

$\sum_{n=0}^(\infty)\frac{1}{(2n+1)*(2n+2)} = \log(2)$

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

$\sum_{n=0}^(\infty)\frac{1}{(2n+1)*2^n} = \sqrt(2) * log(1+\sqrt(2))$

$\sum_{n=0}^(\infty)\frac{(2n + 1)!!}{(2n+1)!} = \sqrt(\e)$

$\sum_{n=0}^(\infty)\frac{(n!)^2}{(2n+1)!} = \frac{2\pi}{3\sqrt(3)}$

$\sum_{n=0}^(\infty)\frac{2^(n+1) * (n!)^2}{(2n+1)!} = \pi$

$\sum_{n=0}^(\infty)\frac{(-1)^n}{2n+1} = \frac{\pi}{4}$

$\sum_{n=0}^(\infty)\frac{\i^(2n+1)}{2n+1} = \frac{\pi\i}{4}$

$\sum_{n=0}^(\infty)\frac{(\i\sqrt(2)-i)^(2n+1)}{2n+1} = \frac{\pi\i}{8}$

$\sum_{n=1}^(\infty)\frac{1}{n^2} = \frac{\pi^2}{6}$

$\sum_{n=1}^(\infty)\frac{1}{n * (n+1)^2} = 2-\frac{\pi^2}{6}$

$\sum_{n=0}^(\infty)\frac{(-1)^n}{(2n+1)^3} = \frac{\pi^3}{32}$

$\sum_{n=0}^(\infty)\frac{1}{n + (2n+1)^2} = \frac{\pi}{6} + \ln(2)$

$\sum_{n=0}^(\infty)\frac{1}{(2n+1)^2} = \frac{\pi^2}{8}$

$\sum_{n=0}^(\infty)\frac{(-1)^n}{(n+1)^2} = \frac{\pi^2}{12}$

$\sum_{n=0}^(\infty)\frac{2^(2n) * B_{2n}}{(2n)!} = \frac{e^2+1}{e^2-1}$

$\sum_{n=0}^(\infty)\frac{(2x)^(2n) * B_{2n}}{(2n)!} = x*\coth(x)$ ,

$0 < |x| < \pi$ .

$\sum_{n=0}^(\infty)\frac{(-1)^n * (2x)^(2n) * B_{2n}}{(2n)!} = x*\cot(x)$ ,

$0 < |x| < \pi$ .

$\sum_{n=1}^(\infty)\frac{2^(2n) * (2^(2n)-1) * x^(2n-1) * B_{2n}}{(2n)!} = \tanh(x)$ ,

$|x| < \pi/2$ .

$\sum_{n=1}^(\infty)\frac{(-1)^(n-1) * 2^(2n) * (2^(2n)-1) * x^(2n-1) * B_{2n}}{(2n)!} = \tan(x)$ ,

$|x| < \pi/2$ .

$sum_{n=0}^(\infty)\frac{1}{(2n+x)^k} = -\frac{\psi^((k-1))(x/2)}{2^k * (k-1)!}$

$\sum_{n=0}^(\infty)\frac{(-1)^n}{(2n+1)^2} = \beta(2)$

$\sum_{n=1}^(\infty)\frac{1}{n+n^2} = 1$

$\sum_{n=0}^(\infty)\frac{x^(2n) * (-1)^n}{(2n)!} = \cos(x)$

$\sum_{n=0}^(\infty)\frac{x^(2n+1) * (-1)^n}{(2n+1)!} = \sin(x)$

$\sum_{n=0}^(\infty)\frac{(\x\i)^(2n+1) - (-\x\i)^(2n+1)}{(2n+1)!} = \frac{2}{\csc(x)}i$

$\sum_{n=0}^(\infty)\frac{x^(2n)}{(2n)!} = \cosh(x)$

$\sum_{n=0}^(\infty)\frac{x^(2n+1)}{(2n+1)!} = \sinh(x)$

$\sum_{n=0}^(\infty)\frac{(2x)^n}{n!} = \frac{2}{1-\tanh(x)}-1$

$\sum_{n=0}^(\infty)\frac{(2x)^n}{n!} = 1-\frac{2}{1-\coth(x)}$

$\sum_{n=0}^(\infty)\frac{x^(2n+1) - (-x)^(2n+1)}{(2n+1)!} = \frac{2}{\csch(x)}$

$\sum_{n=0}^(\infty)\frac{(\frac{sqrt(x^2+1)-1}{x}i)^(2n+1)}{2n+1} = \frac{\tan^-1(x)}{2}i$

$\sum_{n=0}^(\infty)\frac{(\frac{\sqrt(x^2+1)-1}{x})^(2n+1)}{2n+1} = \frac{\sinh^-1(x)}{2}$

$\sum_{n=0}^(\infty)\frac{(\frac{1-sqrt(1-x^2)}{x}i)^(2n+1)}{2n+1} = \frac{\sin^-1(x)}{2}i$

$\sum_{n=0}^(\infty)\frac{(\frac{1-sqrt(1-x^2)}{x})^(2n+1)}{2n+1} = \frac{\tanh^-1(x)}{2}$

$\sum_{n=0}^(\infty)\frac{(sqrt(\frac{x-1}{x+1})i)^(2n+1)}{2n+1} = \frac{\sec^-1(x)}{2}i$

$\sum_{n=0}^(\infty)\frac{sqrt(\frac{x-1}{x+1})^(2n+1)}{2n+1} = \frac{\cosh^-1(x)}{2}$

$\sum_{n=0}^(\infty)\frac{(sqrt(\frac{1-x}{1+x})i)^(2n+1)}{2n+1} = \frac{\cos^-1(x)}{2}i$

$\sum_{n=0}^(\infty)\frac{sqrt(\frac{1-x}{1+x})^(2n+1)}{2n+1} = \frac{\sech^-1(x)}{2}$

$\sum_{n=0}^(\infty)\frac{((1-sqrt(1-\frac{1}{x^2}))\x\i)^(2n+1)}{2n+1} = \frac{\csc^-1(x)}{2}i$

$\sum_{n=0}^(\infty)\frac{((1-sqrt(1-\frac{1}{x^2}))\x)^(2n+1)}{2n+1} = \frac{\coth^-1(x)}{2}$

$\sum_{n=0}^(\infty)\frac{((sqrt(\frac{1}{x^2}+1)-1)\x\i)^(2n+1)}{2n+1} = \frac{\cot^-1(x)}{2}i$

$\sum_{n=0}^(\infty)\frac{((sqrt(\frac{1}{x^2}+1)-1)\x)^(2n+1)}{2n+1} = \frac{\csch^-1(x)}{2}$

$\sum_{n=0}^(\infty)\frac{x^(-2n-1)}{2n+1} = \coth^-1(x)$ for |x| > 1

$\sum_{n=0}^(\infty)\frac{x^\frac{2n}{y}}{(2n)!} = \cosh(x^(1/y))$

$\sum_{n=0}^(\infty)\frac{x^\frac{2n+1}{y}}{(2n+1)!} = \sinh(x^(1/y))$

$\sum_{n=0}^(\infty)\frac{1}{n^2 + (n+1)^2} = \tanh(\frac{\pi}{2}) * \frac{\pi}{2}$

$\sum_{n=0}^(\infty)\frac{(-1)^n}{n+1} = \ln(2)$

$\sum_{n=0}^(\infty)\frac{i^n}{n+1} = \frac{\pi}{4} + \log(\sqrt(2))\i$

$\sum_{n=0}^(\infty)\frac{i^n}{n!} = \cos(1) + \sin(1)\i$

$\sum_{n=0}^(\infty)\frac{i^n}{(2n)!} = \cosh((-1)^(1/4))$

$\sum_{n=1}^(\infty)\frac{1}{n^k * k^n} = Li_k(1/k)$ for |k| > 1

$\sum_{n=0}^(\infty)\frac{i^n}{\e^n} = \e/(\e-\i)$

$\sum_{n=0}^(\infty)\frac{(\frac{x-1}{x+1})^(2n+1)}{2n+1} = \frac{\ln(x)}{2}$

$\sum_{n=0}^\(\infty)\frac{\sum_{k=2}^(m)(\frac{k-1}{k+1})^(2n+1)}{2n+1} = \frac{\ln(m!)}{2}$

$\sum_{n=0}^(\infty)\frac{\prod_{k=0}^(j)(\a_k)^n}{n!} = exp(\prod_{k=0}^(j)\a_k)$

$\sum_{n=0}^(\infty)\frac{(-\frac{1}{4})^n * \cos^(-1)(3)^(2n)}{(2n)!} = \sqrt(2)$

$\sum_{n=0}^(\infty)\frac{\zeta(2n)}{(2n+1)*(2n+2)*2^(2n)} = \frac{-14*\zeta(3)}{8\pi^2}$

$[log(27/16) + \sum_{n=0}^(\infty)\frac{\zeta(2n) - 1}{(2n+1) * (2n+2) * 2^(2n)}] * \frac{-8\pi^2}{14} = \zeta(3)$

$\sum_{n=0}^(\infty)(-\frac{1}{4})^n * (\frac{1}{1+2n} + \frac{2}{1+4n} + \frac{1}{3+4n}) = \pi$

$\sum_{n=1}^(\infty)\frac{\log(n)}{n^2} = -\frac{pi^2}{6} * (\gamma - 12*log(\A) + log(2\pi))$

# Infinite products

$\prod_{p}\frac{1}{1-p^-2} = \frac{\pi^2}{6}$

$\prod_{n=1}^(\infty)\frac{4n^2}{4n^2-1} = \frac{\pi}{2}$

$\prod_{n=0}^(\infty)\exp(\frac{1}{n!}) = \e^\e$

$\prod_{n=1}^(\infty)\frac{4n^2+1}{4n^2-1} = \sinh(\frac{\pi}{2})$

$\prod_{n=1}^(\infty)\frac{4n^2-1}{4n^2+1} = \csch(\frac{\pi}{2})$

$\prod_{n=1}^(\infty)\frac{4n^3}{4n^3 - 3n + 1} = \pi$

$\prod_{n=1}^(\infty)\frac{(2n+1)^2-1}{(2n+1)^2} = \frac{\pi}{4}$

$\prod_{n=1}^(\infty)(1-\frac{1}{4n^2}) = \frac{2}{\pi}$

$\prod_{n=1}^(\infty)(1-\frac{x^2}{\pi^2 * n^2}) = \frac{\sin(x)}{x}$

$\prod_{n=0}^(\infty)(1-\frac{4x^2}{\pi^2 * (2n+1)^2}) = \cos(x)$

$\prod_{n=1}^(\infty)\frac{x*n^2}{x*n^2 - 1} = \frac{\pi}{\sqrt(x)*\sin(\frac{\pi}{\sqrt(x)})}$

$\prod_{n=1}^(\infty)\frac{x*pi^2*n^2}{x*pi^2*n^2 - 1} = \frac{\csc(\frac{1}{\sqrt(x)})}{\sqrt(x)}$

$\prod_{n=1}^(\infty)(1 + \frac{1}{n})^x (1 + \frac{x}{n})^(-1) = \Gamma(x+1)$

$\prod_{n=1}^(\infty)\exp(\frac{1}{n}-2) (\frac{1}{n} + 1)^(2n) = \frac{\exp(2+\gamma)}{2\pi}$

# Continued fractions

$\mathbf{K}(\frac{1}{1}) = φ - 1$

$\mathbf{K}(\frac{1}{2}) = \sqrt(2) - 1$

$\mathbf{K}(\frac{x-1}{2}) = \sqrt(x) - 1$

$\mathbf{K}(\frac{1}{x}) = \frac{\sqrt(x^2+4) - x}{2}$

$\mathbf{K}(\frac{x}{1}) = \frac{\sqrt(4x+1) - 1}{2}$

$\mathbf{K}(\frac{x}{y}) = \frac{\sqrt(y^2 + 4x) - y}{2}$

$\underset{n=0}{\overset{\infty}{\mathbf{K}}}(\frac{(-1)^n}{1}) = φ$

$\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{n}{n}) = \frac{1}{\e-1}$

$\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{1}{2n + 1}) = \frac{2}{\e^2-1}$

$\underset{n=0}{\overset{\infty}{\mathbf{K}}}(\frac{1}{2n + 1}) = \tanh(1)$

$\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{2n}{n-1}) = \coth(1)$

$\underset{n=0}{\overset{\infty}{\mathbf{K}}}(\frac{(2n+1)^2}{6}) = \pi-3$

$\underset{n=0}{\overset{\infty}{\mathbf{K}}}(\frac{(2n+1)^2}{2}) = \frac{4}{\pi}-1$

$\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{n^2}{2n + 1}) = \frac{4}{\pi}-1$

$\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{n^4}{2n + 1}) = \frac{12}{\pi^2}-1$

$\frac{2}{1+\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{n^4}{2n+1})} = \zeta(2)$

$1+\frac{1}{5+\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{-n^6}{(2n+1)*(n^2+n+5)})} = \zeta(3)$

$\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{\floor(\frac{n+1}{2}) * \floor(\frac{n+2}{2})}{1}) = \frac{6}{\pi^2 - 6}-1$

$1+\frac{x}{1 + \underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{-n * x}{x + n + 1})} = \exp(x)$

$\frac{x}{1+\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{n^2 * x^2}{2n+1})} = \tan^-1(x)$

$\frac{1}{1+\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{\frac{x^2}{4n^2 - 2n}}{1-\frac{x^2}{4n^2 - 2n}})} = \cos(x)$

$\frac{x}{1+\underset{n=1}{\overset{\infty}{\mathbf{K}}}(\frac{\frac{x^2}{4n^2 + 2n}}{1-\frac{x^2}{4n^2 + 2n}})} = \sin(x)$

# Nested radicals

$\prod_{n=1}^(\infty)\frac{2}{\underset{1}{\overset{n}{\mathbf{R}}}(2+R)^(1/2)} = \frac{\pi}{2}$

$(1+R)^(1/2) = \sqrt(1+\sqrt(1+\sqrt(1+\...))) = φ$

$(x+R)^(1/2) = \sqrt(x+\sqrt(x+\sqrt(x+\...))) = \frac{1 + \sqrt(4x+1)}{2}$

$(1+R)^(1/3) = (1 + (1 + (1 + \...)^(1/3))^(1/3))^(1/3) = \rho$

$\lim_{n to \infty}((2 - \underset{1}{\overset{n}{\mathbf{R}}}(2 + R)^(1/2)) * 4^(n+1)) = (2 - \sqrt(2 + \sqrt(2 + \sqrt(2 + \...)))) * 4^(n+1) = \pi^2$

$\underset{n=1}{\overset{\infty}{\mathbf{R}}}(1 + n * R)^(1/2) = \sqrt(1 + 1*\sqrt(1 + 2*\sqrt(1 + 3*\...))) = 2$

$\underset{n=x}{\overset{\infty}{\mathbf{R}}}(1 + n * R)^(1/2) = \sqrt(1 + x*\sqrt(1 + (x+1)*\sqrt(1 + (x+2)*\...))) = x+1$

# Limits

$\lim_{n to \infty}(1 + 1/n)^n = \e$

$\lim_{n to \infty}(\frac{n-1}{n})^(-n) = \e$

$\lim_{n to \infty}[prod_{p}^(n) p^(1/n)] = \e$

$\lim_{n to \infty}[prod_{k=1}^(n) p_k^(1/(n \ln n))] = \e$

$\lim_{n to \infty}[\H_n - \ln n] = \gamma$

$\lim_{n to \infty}\frac{\exp(\H_n)}{n} = \exp(\gamma)$

$\lim_{n to \infty}[\zeta((n+1)/n) - n] = \gamma$

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

$\lim_{n to \infty}\frac{n!}{\sqrt(n) * e^(-n) * n^n} = \sqrt(2\pi)$

$\lim_{n to \infty}[sum_{k=0}^(n)\frac{n! * n^(k-n)}{k!*(n-k)!}] = \e$

$\lim_{n to \infty}[(n-1)^(-n) * (n+1)^n] = \e^2$

$\lim_{n to \infty}[n^(1-n) * (n + x)^(n-1)] = \exp(\x)$

$lim_{n to \infty}\frac{2^(4n)*n*(n!)^4}{((2n+1)!)^2} = \frac{\pi}{4}$

$lim_{n to \infty}\frac{2^(4n-1) * (n!)^4}{(2n+1) * ((2n)!)^2} = \frac{\pi}{4}$

$lim_{n to \infty}\frac{2^(4n)*(n!)^3*(n-1)!}{((2n)!)^2} = \pi$

$lim_{n to \infty}\frac{2^(4n)*(n!)^3*(n+1)!}{((2n+1)!)^2} = \frac{\pi}{4}$

$lim_{n to \infty}\frac{4^n * (n!)^2}{(2n-1)!! * (2n+1)!!} = \frac{\pi}{2}$

$lim_{n to \infty}\frac{2^(8n)*n*(n+1)!*(n!)^7}{((2n+1)!)^4} = \frac{\pi^2}{16}$

$lim_{n to \infty}\frac{2^(8n+1)*(n+1)^4*(2n^2 + 2n + 1) * (n!)^8}{((2n+2)!)^4} = \frac{\pi^2}{8}$

$lim_{n to \infty}[n^2/4 - \frac{6n^2+6n+1}{12}*log(n) + sum_{k=1}^(n)k \ln k] = \log(\A)$

# Asymptotics

$sqrt(2\pi\n) * (n/e)^n ~~ n!$

$2^n * sqrt(2 \pi n) * (n/e)^n ~~ (2n)!!$

$2^(n+1/2) * (n/e)^n ~~ (2n-1)!!$

$2^(n+3/2) * ((n+1)/e)^(n+1) ~~ (2n+1)!!$

$n * (log(n) + log(2) - 1) + log(2 \pi n)/2 ~~ log((2n)!!)$

$n*log(n) + (n + 1/2)*log(2) - n ~~ log((2n-1)!!)$

$(n+1)*log(n+1) + (n + 3/2)*log(2) - n - 1 ~~ log((2n+1)!!)$

$\frac{2}{(2\pi)^n} * n! ~~ |B_n|$

$\sqrt(8\pi\n) * (\frac{n}{2\pi\e})^n ~~ |B_n|$

$\log(\sqrt(8\pi\n)) + n * (log(n) - log(2\pi) - 1) ~~ log(|B_n|)$

$-\frac{n^2}{4} + \frac{6n^2 + 6n + 1}{12} * log(n) + log(\A) ~~ \sum_{k=1}^(n) k \ln k$

$\sqrt(2\pi\n) * \exp(n*log(n) - n) * (\log(n) + \gamma) ~~ n!*\sum_{k=1}^(n) \frac{1}{k}$

$\log(\sqrt(2\pi\n)) + n*log(n) + log(log(n) + \gamma) - n ~~ log(n!*\sum_{k=1}^(n) \frac{1}{k})$

$\sqrt(2\pi) * (\frac{12n^2 + 24n + 12 + 1}{12\e\n + 12\e})^(n+1) * \frac{1}{\sqrt(n+1)} ~~ n!$

$1/2 * (\log(\frac{2\pi}{n+1}) - 2*(n+1) * \log(\frac{12\e\n + 12\e}{12n^2+24n+12+1})) ~~ \log(n!)$

$1/6 * (\log(8n^3 + 4n^2 + n + 1/30) - 6n + 6n*log(n) + 3\log(\pi)) ~~ \log(n!)$

$lim_{k to \infty}[\sqrt(2\pi) * (\frac{12n^2 + 24kn + 12k^2 + 1}{12\e\n + 12\e\k})^(n+k) * \frac{n!}{(n+k-1)!} * \frac{1}{\sqrt(n+k)}] ~~ n!$

Mathematics and computer science