Series that converge to $\pi$ quickly I know the series, $4-{4\over3}+{4\over5}-{4\over7}...$ converges to $\pi$ but I have heard many people say that while this is a classic example, there are series that converge much faster.  Does anyone know of any?
 A: I think you may find interesting to browse the webpage of Jon Borwein, which I would call the standard reference for your question. In particular, take a look at the latest version of his talk on "The life of pi" (and its references!), which includes many of the fast converging algorithms and series used in practice for high precision computations of $\pi$, such as the one from this Summer.
A: Just to give people an idea on convergence rates, here is a plot of $-\log_{10}\left|\frac{S_n-\pi}{\pi}\right|$ versus $n$ , where $S_n$ is the nth partial sum of the series in question, for three of the series featured in the answers to this question (note the vertical scale):

The three series are, from top to bottom, $\arctan(1)$ (the series mentioned by the OP), $2\arcsin\left(\sqrt{\frac12}\right)$ (the series mentioned by yjj in his answer), and the series by Ramanujan I mentioned in the comments (I didn't include the series by the Chudnovsky brothers, since that converges even faster than the Ramanujan series, and that makes for boring plots).
A: Here is a really nice one due to Simon Plouffe. There are many similar examples in his linked paper.
$$\pi = 72\sum_{n=1}^\infty \frac{1}{n(e^{n\pi} - 1)} - 
96\sum_{n=1}^\infty \frac{1}{n(e^{2n\pi} - 1)} +
24\sum_{n=1}^\infty \frac{1}{n(e^{4n\pi} - 1)} .$$
What I like about it is that I can see at a glance that the series converge rapidly without having to make some mental estimate of the size of factorials.
A: You should take a look at the paper: Some New Formulas for π by Gert Almkvist, Christian Krattenthaler, and Joakim Petersson, Experiment. Math. Volume 12, Number 4 (2003), 441-456.
A: Your series may be written as
$$\frac{\pi}{4}=\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)$$
Its truncation approximations improve if the zero relation (http://oeis.org/A176563)
$$0=\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)$$
is added to obtain
$$\frac{\pi}{4}=\sum_{k=0}^{\infty}\left(\frac{2}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+4}\right)$$
$$=\frac{3}{4}\sum_{k=0}^{\infty}\frac{1}{(4k+1)(2k+1)(k+1)}$$
(Lehmer, http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf, http://oeis.org/A079588)
Although this is the slowest series in all answers, it illustrates how an absolutely convergent series of unit fractions for $\frac{\pi}{3}$ may be obtained by summing up two conditionally convergent series that have been regrouped.
This simple series also explains Why is $\pi$ so close to $3$? by taking the first term out of the summation.
A: The series $$ \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^n = \frac{\pi}{2}$$ converges quickly. Here $!!$ is the double factorial defined by $0!! = 1!! = 1$ and $n!! = n (n-2)!!$
This is series is not too hard to derive. Start by defining $$f(t) = \sum _{n=0}^{\infty } \frac{(-1)^n}{(2n+1)}t^n.$$ Note that $f(1) = \pi/4$ is the series you referenced. Now we take what is called the Euler Transform of the series which gives us $$ \left(\frac{1}{1-t}\right)f\left(\frac{t}{1-t}\right) = \sum _{n=0}^{\infty } \left(\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)}\right)t^n.$$
Now $$\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)} = \frac{(2n)!!}{(2n+1)!!}$$ for hints on how to prove this identity see Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$. Now put $t = 1/2$ and the identity follows. Showing the error term for the nth partial sum is less than $(1/2)^n$ is not too difficult.
A: The BBP formula is another nice one:
$$
  \pi = \sum_{k=0}^\infty \left[
  \frac{1}{16^k} \! \left(
  \frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}
  \right) \right]
$$
It can be used to compute the $n$th hexadecimal digit of $\pi$ without computing the preceding $n{-}1$ digits.
A: Not the fastest but much faster than a lot of the "simple" formulas you may have heard:
$\pi=\displaystyle\sum^{\infty}_{n=0}{\frac{n!\left(2n\right)!\left(25n-3\right)}{2^{n-1}\left(3n\right)!}}$
Produces a digit or more of $\pi$ per term.
A: Maybe this one discovered by Newton.
A: This one is possibly the fastest converging formula with rational terms $a,b,r$ of the form $\pi=c\sum^{\infty}_{n=0}(a+bn)r^n$ , which adds a little over 2 digits per term, or more exactly  $\log (324)$:
$\frac{\sqrt{3}}{60}\displaystyle\sum^{\infty}_{n=0} \frac{(-1)^n(2n)!(130n+109)}{(7/6)_{n}(11/6)_{n}6^{4n}}$
testing it
[2] source: https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3919892
A: The convergence can be arbitrary fast unless you don't specify what kind of series you are looking. Let $k$ be a positive integer, $a_n=\pi/k$ if $n\leq k$ and zero elsewhere. Then $\sum_{n=1}^\infty a_n$ converges to $\pi$ after $k$ summands.
