# From Ramanujan

This is a collection of formulas discovered by the Indian Mathematician Srinivasa Ramanujan (1887-1920) that appears in the book “A Passion for Mathematics” by Clifford A Pickover. $1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\ldots=\cfrac{1}{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ldots}}}}=\sqrt{\frac{\pi e}{2}}$ $\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum_{k=0}^{\infty}\frac{(4k)!(1103+26390k)}{(k!)^4396^{4k}}$ $\frac{1}{\pi}=\sqrt{8}\sum_{n=0}^\infty\frac{(1103+26390n)(2n-1)!!(4n-1)!!}{99^{4n+2}32^n(n!)^3}$ $\frac{1}{e-1}=\cfrac{1}{1+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\ldots}}}}$ $\pi \approx \frac{99^2}{2206\sqrt{2}}\text{(correct to 8 places)}$ $\frac{e^\pi-1}{e^\pi+1}=\cfrac{\pi}{2+\cfrac{\pi^2}{6+\cfrac{\pi^2}{10+\cfrac{\pi^2}{14+\ldots}}}}$ $\frac{e^x-1}{e^x+1}=\cfrac{x}{2+\cfrac{x^2}{6+\cfrac{x^2}{10+\cfrac{x^2}{14+\ldots}}}}$ $\prod_p^\infty \left( \frac{p^2+1}{p^2-1} = \frac{5}{2} \right)$

Here, the infinite product symbolized by $\prod$, steps through every prime number. $\frac{1}{2}=\sqrt{1-\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{1-\ldots}}}}}$ $1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1\cdot3}{2\cdot4}\right)^3-13\left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right)^3+\ldots=\frac{2}{\pi}$ $\int_0^a e^{-x^2}dx=\frac{1}{2}\pi^{1/2}-\cfrac{e^{-a^2}}{2a+\cfrac{1}{a+\cfrac{2}{2a+\cfrac{3}{a+\cfrac{4}{2a+\ldots}}}}}$ $\frac{1}{1+\cfrac{e^{-2\pi}}{1+\cfrac{e^{-4\pi}}{1+\ldots}}}=\left(\sqrt{\frac{5+\sqrt{5}}{2}}-\frac{\sqrt{5}+1}{2}\right)\cdot e\frac{2}{5^\pi}$

Note that the term in parentheses can be reduced to $\sqrt{2+\phi}-\phi$, where $\phi$ is the golden ratio. $\ln x+\gamma=\sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}{k!k}=x-\frac{x^2}{4}+\frac{x^3}{18}-\frac{x^4}{96}+\ldots$
Where $\gamma=0.5772157\ldots$ is the Euler-Mascheroni constant. $-2=\sqrt{-6+\sqrt{-6+\sqrt{-6+\sqrt{-6+\ldots}}}}$ $3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\ldots}}}}$ $10-\pi^2=\sum_{k=1}^\infty \frac{1}{k^3(k+1)^3}=\frac{1}{1^3\cdot2^3}+\frac{1}{2^3\cdot3^3}+\frac{1}{3^3\cdot4^3}+\ldots$ $\ln 2=\frac{1}{2}+\sum_{k=1}^\infty \frac{1}{(2k)^3-2k}=\frac{1}{2}+\frac{1}{2^3-2}+\frac{1}{4^3-4}+\frac{1}{6^3-6}+\ldots$ $\frac{\pi^2}{6}-3\ln \left(\frac{\sqrt{5}+1}{2}\right)^2=\sum_{k=0}^\infty \frac{(-1)^k(k!)^2}{(2k)!(2k+1)^2}=1-\frac{1}{2!\cdot3^2}+\frac{(2!)^2}{4!\cdot5^2}-\frac{(3!)^2}{6!\cdot7^2}+\ldots$ $\frac{x}{2x-1}=1-\frac{x-1}{x+1}+\frac{(x-1)(x-2)}{(x+1)(x+2)}-\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}+\ldots$

The following formulas used Pochhammer notation. In particular, $(x)_n or (x)_{Poch(n)}=x\times(x+1)\times(x+2)\times\ldots\times (x+n-1)$

The subscript value effectively tells you how many terms to include in the product. $\pi=4\left( \sum_{n=0}^{\infty} \frac{(6n+1)\left(\frac{1}{2}\right)_{Poch(n)}^3}{4^n(n!)^3} \right)^{-1}$ $\pi=\frac{27}{4}\left(\sum_{n=0}^\infty \frac{(15n+2)\left(\frac{1}{2}\right)_{Poch(n)}\left(\frac{1}{3}\right)_{Poch(n)}\left(\frac{2}{3}\right)_{Poch(n)}}{(n!)^3}\left(\frac{2}{27}\right)^n\right)^{-1}$ $\pi=\frac{5\sqrt{5}}{2\sqrt{3}}\left(\sum_{n=0}^\infty \frac{(11n+1)\left(\frac{1}{2}\right)_{Poch(n)}\left(\frac{1}{6}\right)_{Poch(n)}\left(\frac{5}{6}\right)_{Poch(n)}}{(n!)^3}\left(\frac{4}{125}\right)^n\right)^{-1}$