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.
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 $1/\pi$.
This is from the Chudnovsky algorithm. See the associated Wikipedia entry for more information.
