The following formula for $\pi$ was discovered by Ramanujan:
$$\frac1{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\!$$
Does anyone know how it works, or what the motivation for it is?
Ramanujan’s $\pi$ Formula – Motivation and Explanation
approximationcalculuspisequences-and-series
Related Solutions
It is easily seen that we can write the formula as $$\frac{1}{\pi} = \frac{2\sqrt{2}}{9801}\sum_{n = 0}^{\infty}\frac{(4n)!(1103 + 26390n)}{4^{4n}(n!)^{4}99^{4n}} = A\sum_{n = 0}^{\infty}B_{n}C_{n}$$ where $$A = \frac{2\sqrt{2}}{9801},\, B_{n} = \frac{(4n)!(1103 + 26390n)}{4^{4n}(n!)^{4}},\,C_{n} = \frac{1}{99^{4n}}$$ Now if we take a close look at $A, B_{n}, C_{n}$ then we note that $A$ is a constant and hence it does not have any impact on the rate of convergence of the series.
The term $B_{n}$ consists of a factorial part and a linear part. The factorial part $(4n)!/4^{4n}(n!)^{4}$ seems to remain almost constant when $n$ increases one by one. Clearly if we further write $B_{n} = D_{n}E_{n}$ where $D_{n} = (4n)!/4^{4n}(n!)^{4}$ and $E_{n} = (1103 + 26390n)$ then we can see that $$\begin{aligned}\frac{D_{n + 1}}{D_{n}} &= \frac{(4n + 4)!}{4^{4n + 4}((n + 1)!)^{4}}\cdot\frac{4^{4n}(n!)^{4}}{(4n)!}\\ &= \frac{(4n + 1)(4n + 2)(4n + 3)(4n + 4)}{4^{4}(n + 1)^{4}}\\ &= \frac{(4n + 1)(4n + 2)(4n + 3)}{(4n + 4)(4n + 4)(4n + 4)}\end{aligned}$$ so that $D_{n + 1}/D_{n}$ is always less than $1$ and tends to $1$ as $n \to \infty$. The linear term $E_{n}$ on the other hand adds the value $26390$ extra in numerator in each successive term and the ratio $E_{n + 1}/E_{n}$ is always greater than $1$ but tends to $1$ as $n \to \infty$.
It follows that the factor $B_{n} = D_{n}E_{n}$ remains roughly constant as $n$ increases one by one. The real change happens because of the factor $C_{n} = 1/99^{4n}$ and we can see that $99^{4n} \approx 100^{4n} = 10^{8n}$ so that $C_{n} \approx (1/10^{8})^{n}$ This gives eight decimal zeroes as $n$ increases one by one. So we can see that it is the term $C_{n}$ which is the key behind getting eight decimal digits per term.
Update: On an unrelated note, this is one of the best results given by Ramanujan which is exceedingly difficult to prove and to date there is no proof available within the limits of hand calculation. Also important is to note that this is an early work of Ramanujan which was done in India before he got in touch with the British mathematician G. H. Hardy. In case you are interested in the theory behind this formula you can refer these posts here.
Proposition 1 :$$\color{blue}{\displaystyle \sum\limits_{k=1}^\infty \dfrac{(-1)^{k-1}}{k}\;\zeta_H(k,a)x^k \; =\; \ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\tag1}$$
Proof :
$$\begin{align*} & \displaystyle \sum\limits_{k=0}^{\infty}(-x)^{k}\zeta_H(k+1,a)\\ & \displaystyle = \sum\limits_{k,n=0}^{\infty} \dfrac{(-x)^k}{(n+a)^{k+1}}\\ & \displaystyle = \sum\limits_{n=0}^\infty \dfrac{1}{n+a}\sum\limits_{k=0}^\infty \left(\dfrac{-x}{n+a}\right)^k \\ & \displaystyle = \sum\limits_{n=0}^\infty \dfrac{1}{n+a+x} \\ & \displaystyle = -\psi(a+x)\end{align*}$$
Now integrating,
$$\displaystyle \sum\limits_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\zeta_H(k,a)x^k =\ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\\ $$
Proposition 2 :$$\color{blue}{\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x^2}{(n+a)^2}\right)\;=\; \dfrac{\Gamma^2 (a)}{\Gamma(a+ix)\Gamma(a-ix)}\tag 2} $$
Proof :
It is sufficient to evaluate the series,
$$\begin{align*} & \displaystyle\sum\limits_{n=0}^{\infty}\ln\left(1+\dfrac{x}{n+a}\right) \\ & = \displaystyle \sum\limits_{k=1}^\infty\sum\limits_{n=0}^\infty \dfrac{(-1)^{k-1}}{k}x^k \dfrac{1}{(n+a)^k} \\ & =\displaystyle \sum\limits_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\zeta_H(k,a)x^k \\ & =\displaystyle \ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\end{align*}$$
$$\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x}{n+a}\right) = \dfrac{\Gamma(a)}{\Gamma(a+x)}$$
Therefore,
$$\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x^2}{(n+a)^2}\right)\;=\; \dfrac{\Gamma^2 (a)}{\Gamma(a+ix)\Gamma(a-ix)}$$
Proposition 3 : If $\; \displaystyle F(s)=\int\limits_0^\infty x^{s-1}f(x)\; dx\; $ then $$\color{blue}{\displaystyle \int\limits_{-\infty}^{\infty} |F(ix)|^2 \; dx \;= \; 2\pi\int\limits_0^\infty \dfrac{|f(x)|^2}{x}\; dx\tag 3} $$
Proof :
$$\displaystyle F(it)=\int\limits_0^\infty x^{it}\dfrac{f(x)}{x}\; dx$$
Set $x=e^y$ ,
$$\displaystyle F(it)=\int\limits_0^\infty e^{ixt}f(e^x)\; dx$$
Now by properties of Fourier Transform,
$$\begin{align*} & \displaystyle f(e^t)=\int\limits_{-\infty}^\infty g(x)e^{-ixt}\; dx \\ & \displaystyle g(t)=\dfrac{1}{2\pi}\int\limits_{-\infty}^\infty f(e^x)e^{ixt}\; dx\\\end{align*}$$
$$\begin{align*}\displaystyle F(it) & =2\pi g(t) \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 4\pi^2 \int\limits_{-\infty}^\infty |g(t)|^2\; dt \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty g(t) \int\limits_{-\infty}^\infty e^{ixt}f(e^x)\; dx\; dt \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty f(e^x) \int\limits_{-\infty}^\infty e^{ixt}g(t)\; dt\; dx \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty f(e^x)\overline{f(e^x)}\; dx =2\pi\int\limits_{-\infty}^\infty |f(e^x)|^2\; dx\end{align*}$$
Now by setting $e^x=t$ we get our result,
$$\displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt = 2\pi\int\limits_{-\infty}^\infty \dfrac{|f(t)|^2}{t}\; dt$$
Main Problem: $$\color{blue}{\displaystyle \int\limits_{0}^{\infty}\frac {1+\dfrac {x^2}{(b+1)^2}}{1+\dfrac {x^2}{a^2}}\dfrac {1+\dfrac {x^2}{(b+2)^2}}{1+\dfrac {x^2}{(a+1)^2}}\dfrac {1+\dfrac {x^2}{(b+3)^2}}{1+\dfrac {x^2}{(a+2)^2}}\cdots \, dx=\dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}} $$
Proof : If we denote the integral by $I$ then using $(2)$ it can be rewritten as,
$$\displaystyle I = \dfrac{\Gamma^2 (b+1)}{\Gamma^2 (a)}\dfrac{1}{2} \int\limits_{-\infty}^\infty \dfrac{|\Gamma(a+ix)|^2}{|\Gamma(b+1+ix)|^2}\; dx$$
Now by defining $\displaystyle h(x)=\dfrac{x^a (1-x)^{b-a}}{\Gamma(b-a+1)}$ for $x\in[0,1]$ and $0$ for $\forall x\notin[0,1]$ we can conclude that $\displaystyle F(s)=M[h(x)]=\dfrac{\Gamma(s+a)}{\Gamma(s+b+1)}$ and from $(3)$ it follows that,
$$\displaystyle I = \dfrac{\Gamma^2 (b+1)}{\Gamma^2 (a)}\dfrac{1}{2} \int\limits_{0}^1 \dfrac{|h(x)|^2}{x}\; dx = \dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}$$
where last line follows from the Duplication formula , and we are done !
$$\large \color{red}{\color{blue}{\boxed{\mathfrak{PROVED}}}} $$
Best Answer
Here's an easy introduction to the basics, "Pi Formulas and the Monster Group".
http://sites.google.com/site/tpiezas/0013
Update: Just to make this more intriguing, define the fundamental unit $U_{29} = \frac{5+\sqrt{29}}{2}$ and fundamental solutions to Pell equations,
$$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$
$$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$
$$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$
then we can see those integers all over the formula as,
$$\frac{1}{\pi} =\frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k}$$
See also this MO post.