TITLE: Ramanujan's approximation for $\pi$
QUESTION [27 upvotes]: 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?

REPLY [27 votes]: 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.