Complicated series converges to $\pi$.

Question

How do I get this result? $$\frac {426880 \sqrt {10005}}{\large \sum_{k = 0}^{\infty}\frac {(6k)!(545140134k + 13591409)}{(k!)^3 (3k)! (-640320)^{3k}}} = \pi$$

It seems formidable.

Context: I came across this when reading a book (in Chinese) on sequences for high school students where the author started to introduce infinite sequences. He listed a few other famous results and then this, as an illustration that this kind of series can be really complicated.

Answer 1

This is from the Chudnovsky algorithm. See the associated Wikipedia entry for more information.

Answer 2

This series was obtained by Chudnovsky brothers through the theory of modular forms. To describe their formula let us define functions $P, Q, R$ via the equations \begin{align} P(q) &= 1 - 24\left(\frac{q}{1 - q} + \frac{2q^{2}}{1 - q^{2}} + \frac{3q^{3}}{1 - q^{3}} + \cdots\right)\notag\\ Q(q) &= 1 + 240\left(\frac{q}{1 - q} + \frac{2^{3}q^{2}}{1 - q^{2}} + \frac{3^{3}q^{3}}{1 - q^{3}} + \cdots\right)\notag\\ R(q) &= 1 - 504\left(\frac{q}{1 - q} + \frac{2^{5}q^{2}}{1 - q^{2}} + \frac{3^{5}q^{3}}{1 - q^{3}} + \cdots\right)\notag \end{align} where $q$ is a real number with $|q| < 1$ (so that the series above are convergent).

Let $n > 3$ be an integer and we write \begin{align} P_{n} &= P(-e^{-\pi\sqrt{n}}),\,\, Q_{n} = Q(-e^{-\pi\sqrt{n}}),\,\, R_{n} = R(-e^{-\pi\sqrt{n}})\notag\\ j_{n} &= 1728\frac{Q_{n}^{3}}{Q_{n}^{3} - R_{n}^{2}}\notag\\ b_{n} &= \sqrt{n(1728 - j_{n})}\notag\\ a_{n} &= \frac{b_{n}}{6}\left\{1 - \frac{Q_{n}}{R_{n}}\left(P_{n} - \frac{6}{\pi\sqrt{n}}\right)\right\}\notag \end{align} then Chudnovsky brothers establish the following general series for $1/\pi$ $$\boxed{{\displaystyle \frac{1}{\pi} = \frac{1}{\sqrt{-j_{n}}}\sum_{m = 0}^{\infty}\frac{(6m)!}{(3m)!(m!)^{3}}\frac{a_{n} + mb_{n}}{j_{n}^{m}}}}\tag{1}$$ The series in question is obtained by putting $n = 163$. Obtaining the actual numbers given in the series (like $13591409$) from the value of $n = 163$ is very very difficult via hand calculation.

I have shown in my blog post that the above general series $(1)$ can be obtained through Ramanujan's theory described in Ramanujan's most famous paper Modular Equations and Approximations to $\pi$ without using the deep and difficult theory of modular forms. Ramanujan explicitly did not obtain the formula $(1)$ but indicated a general method to obtain series for $1/\pi$ and the series given by Chudnovsky can also be obtained via this method. See my blog post for more details.

Also see one of my answers regarding Ramanujan's series for $1/\pi$.