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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?

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Here is a relevant reference:… – Jonas Meyer Dec 13 '10 at 1:03
A closely related question:… – Jonas Meyer Dec 13 '10 at 1:05
That would be Ramanujan's $$\frac{1}{\pi} = \frac{2 \sqrt 2}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$$ or the Chudnovsky brothers' $$\frac{426880 \sqrt{10005}}{\pi} = \sum_{k=0}^\infty \frac{(6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 (-640320)^{3k}}$$ – J. M. Dec 13 '10 at 1:15
As an additional note: when using the series for the arctangent, the series converges more slowly as one is nearing the convergence boundary $|z|=1$ ; the reason Machin's formulae work well is that it expresses $\pi$ as sums of arctangent series with arguments near the point of expansion $z=0$. – J. M. Dec 13 '10 at 1:20
@J.M. I think your last comment, if expanded a bit, can be a good answer to the question. – Willie Wong Dec 13 '10 at 1:38
up vote 9 down vote accepted

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.

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That's not true: The first n-1 elements in the series do impact the n-th digit, since each element is just a number with an infinite hexadecimal expansion starting around the k-th digit. it impacts the other digits a lot. – Ekuurh Sep 17 '14 at 9:46
@Ekuurh: You're right, the formulation "can be used" might be a bit misleading. The formula can be used for what I claim, but with increasing $n$ the computing time increases like $n\ln n$. – Hendrik Vogt Sep 19 '14 at 6:58

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. 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.

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You know, you could've used $\arctan$ instead of $f$, and it'd be a bit clearer... :) – J. M. Dec 13 '10 at 9:05
In hypergeometric form, the first series is ${}_2 F_1\left(1,1;\frac32;\frac12\right)=2\arcsin\left(\sqrt{\frac12}\right)$ . – J. M. Dec 13 '10 at 9:14

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.

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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.

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...but there's a $\pi$ in the individual terms... :D – J. M. Dec 13 '10 at 11:24
@J.M. Agreed, this does detract from it somewhat, but it still impresses me nevertheless. – Derek Jennings Dec 13 '10 at 11:30

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):

partial sum plots

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).

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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.

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...I'm missing the point of this answer, apparently. – J. M. Dec 13 '10 at 15:21
-1: The whole point of the question is to be able to compute a reasonable approximation to $\pi$ quickly. If your terms involve $\pi$, you are stuck in a loop... – Aryabhata Dec 13 '10 at 17:22
@Jaska: Any real number $x$ can be represented as a "sum" of just one real number, namely $x=x$. If you don't like this "sum", say $x=x/2+x/2$. This is true, but not helpful at all. (By the way, I didn't vote down.) – Hendrik Vogt Dec 13 '10 at 17:43
@Hendrik Vogt: That is true. And in my opinion @Joe should tell first what kind of real numbers he wants to use in the series. There is also $\pi$ in Derek Jennings's post so I'm wondering why my series is worse than the one Jennings gave. – Jaska Dec 13 '10 at 18:11
@Jaska: Derek cited an interesting series, which is, however, not helpful, as it contains $\pi$ (as J.M. already remarked). No offence meant at all: I wouldn't say your finite sum is worse, but I'd regard it neither interesting nor helpful. – Hendrik Vogt Dec 13 '10 at 18:23

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.

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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 ( $$0=\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)$$

is added to obtain $$\frac{\pi}{4}=\sum_{k=0}^{\infty}\left(\frac{2}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+4}\right)$$ $$=\frac{3}{4}\sum_{k=0}^{\infty}\frac{1}{(4k+1)(2k+1)(k+1)}$$


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.

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protected by t.b. Jul 30 '11 at 10:53

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