Ramanujan's approximation for $\pi$

Question

In 1910, Srinivasa Ramanujan found several rapidly converging infinite series of $\pi$, such as $$ \frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}. $$

Wikipedia says this formula computes a further eight decimal places of $\pi$ with each term in the series. There are also generalizations called Ramanujan–Sato series.

I tested it with Maple and really each term gives more eight right places. Anyone could tell me, how could each step give more eight correct digits?

Answer 1

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.

Answer 2

Using Stirling's formula, $$ \frac{(4k)!}{(k!)^4}\sim\frac{4^{4k}}{\sqrt{2\pi^3k^3}} $$ Thus, as $k\to\infty$, the ratio of terms in the sum is $$ \begin{align} \frac{a_{k+1}}{a_k} &\sim\frac{4^4}{396^4}\frac{k^{3/2}}{(k+1)^{3/2}}\frac{27493+26390k}{1103+26390k}\\ &\to\frac{4^4}{396^4}\\ &=\frac1{99^4}\\ &=\frac1{96059601}\\[6pt] &\approx10^{-8} \end{align} $$ Thus, each term is approximately $10^{-8}$ of the term before it. Thus, the convergence of about $8$ digits per term.