TITLE: How do you calculate the decimal expansion of an irrational number?
QUESTION [32 upvotes]: Just curious, how do you calculate an irrational number? Take $\pi$ for example. Computers have calculated $\pi$ to the millionth digit and beyond. What formula/method do they use to figure this out? How does it compare to other irrational numbers such as $\varphi$ or $e$?

REPLY [5 votes]: An example not yet given.
$\zeta(3)=1.20205690315959428539973816151144$
$\qquad\qquad 999076498629234049...$ (here)
The number $$\zeta (3)=\sum_{n=1}^\infty \frac{1}{n^3} \tag{1}$$ is called Apéry's constant, because its irrationality was first proved by Roger Apéry. The following series, which converges to $\zeta (3)$ faster than $(1)$, can be used to compute it
$$\zeta (3)=\frac{5}{2}\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}}.\tag{2}$$
For the same purpose we can use the continued fraction expansion for $\zeta (3)$, which is
$$\zeta \left( 3\right) =\dfrac{6}{5-\dfrac{1}{117-\dfrac{64}{535-...-\dfrac{n^{6}}{34n^{3}+51n^{2}+27n+5}}}}.\tag{3}$$
Another possibility is to use the following limit
$$\begin{equation*}
\zeta (3)=\lim_{n\rightarrow \infty }\frac{a_{n}}{b_{n}},
\end{equation*}\tag{4}$$
where
$$\begin{equation*}
a_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}c_{n,k},
\end{equation*}\tag{5}$$
$$\begin{equation*}
b_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2},
\end{equation*}\tag{6}$$
and
$$\begin{equation*}
c_{n,k}=\sum_{m=1}^{n}\frac{1}{m^{3}}+\sum_{m=1}^{k}\frac{\left( -1\right)
^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\quad k\leq n.
\end{equation*}\tag{7}$$
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References
Apéry, Roger (1979), Irrationalité de $\zeta 2$ et $\zeta 3$, Astérisque 61: 11–13
Alfred van der Poorten (1979), A proof that Euler missed..., The Mathematical Intelligencer 1 (4): 195–203