# Examples of series approximating $\pi$

The first time I saw this serie is in an article titled “Examples of series approximating $\pi$”. It was said that the most beautiful formula among a lot is this:

$$\pi=\frac{9801}{2\sqrt{2}\sum_{n=0}^{+\infty}\frac{(4n)!}{(n!)^{4}}\frac{1103+26390n}{396^{4n}}}$$ by Ramanujan.

My question is what makes this formula so beautiful?

• It is inscrutable! That is, how did Ramanujan come up with something like this? – Stephen Montgomery-Smith Nov 16 '13 at 22:27
• The first term in the series is within 0.0000025% of $\pi$. – David H Nov 16 '13 at 22:33
• Is there a proof that this formula's limit is $\pi$? I heard that Ramanujan liked to write down correct formulas without proofs. – Leif Sabellek Nov 16 '13 at 22:56
• @LeifSabellek yeah, proposed in $1910$ and proved in $1985$ by Borwein brothers. – Mohamez Nov 17 '13 at 14:09

## 2 Answers

"Beauty" could mean the fact that there is no reason why the combination of numbers put forth should ever be seen to have anything to do with $\pi$ by mere mortals. Or the fact that the series converges so damn quickly.

• I don't think it converges that fast, does it? It looks like that summand is approximately a geometric series. – Stephen Montgomery-Smith Nov 16 '13 at 22:52
• @StephenMontgomery-Smith: it does...but the $n$th term is approx $C n^{-1/2} (99)^{-4 n}$ for large $n$, where $C=2 (2 \pi)^{-3/2} 26390$. Each term is almost $10^8$ smaller than the previous one. To me, that's pretty good. – Ron Gordon Nov 16 '13 at 23:00
• Fair enough. But my idea of rapid convergence is where each extra term doubles the number of digits. I think Ramanujun has an iterative method that does that. – Stephen Montgomery-Smith Nov 16 '13 at 23:57

The following Wikipedia articles and paragraphs are relevant to your question:

• It seems those links are at best loosely related to OP's question. – R R Nov 17 '13 at 11:28
• Why ? In what way ? I don't understand. – Lucian Nov 17 '13 at 12:19