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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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    @NotSuper: The Chudnovskys' formula. Compare the decay rate of the terms of both series...2011-07-30

10 Answers 10

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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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    This is like saying that $\pi=\sum_{i=1}^1 \pi$ is a rapidly converging series. It might be true but it's not useful at all. It's only correct by a technicality.2017-10-14
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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.

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    In hypergeometric form, the first series is ${}_2 F_1\left(1,1;\frac32;\frac12\right)=2\arcsin\left(\sqrt{\frac12}\right)$ .2010-12-13
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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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    Is there a version of that formula that has a 1/10^k term instead of a 1/16^k one?2017-03-14
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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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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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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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    @J.M. Agreed, this does detract from it somewhat, but it still impresses me nevertheless.2010-12-13
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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 (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.

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Here's a formula which I found embedded in an old C program. I don't know where this comes from, but it converges to Pi very quickly, about 16 correct digits in just 22 iterations:

$\pi = \sum_{i=0}^{\infty}{ \frac{6(\prod{2j-1})} {(\prod{2j})(2i+1)(2^{2i+1})}}$

(The products are from 1 to i, so that for i=0 the products are empty, essentially 1/1. For i=1, the products are 1/2. For i=2, the products are (1*3)/(2*4). For i=3, the products are (1*3*5)/(2*4*6). Etc, ad infinitum.)

I have no idea of the provenance of that formula, but on running the C program, it produces:

Index =    0    Sum = 3.000000000000000 Index =    1    Sum = 3.125000000000000 Index =    2    Sum = 3.139062500000000 Index =    3    Sum = 3.141155133928572 Index =    4    Sum = 3.141511172340030 Index =    5    Sum = 3.141576715774867 Index =    6    Sum = 3.141589425319122 Index =    7    Sum = 3.141591982358383 Index =    8    Sum = 3.141592511157862 Index =    9    Sum = 3.141592622870617 Index =   10    Sum = 3.141592646875561 Index =   11    Sum = 3.141592652105887 Index =   12    Sum = 3.141592653258738 Index =   13    Sum = 3.141592653515338 Index =   14    Sum = 3.141592653572930 Index =   15    Sum = 3.141592653585950 Index =   16    Sum = 3.141592653588912 Index =   17    Sum = 3.141592653589590 Index =   18    Sum = 3.141592653589746 Index =   19    Sum = 3.141592653589782 Index =   20    Sum = 3.141592653589790 Index =   21    Sum = 3.141592653589792 Index =   22    Sum = 3.141592653589793 

Which is 16 correct significant figures in just 22 iterations, which is actually pretty darn fast. Many serieses which converge to Pi do so with infuriating slowness, requiring 1000 iterations to get 3.1429384 which wrong after the first 3 digits. But not THIS formula! It generates almost as many good digits as iterations.

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    A quick search finds some reasonably simple series for arcsin though I have only found ones that show a few terms and no general formula for the terms. http://www.efunda.com/math/taylor_series/inverse_trig.cfm What I did long ago was calculate the first few derivatives by hand, guessed a pattern, and proved that pattern by induction. My program was able to calculate one term by a few operations on the previous one. The hard part was developing the routines for working with many decimal places. No off the shelf libraries in those days.2018-05-31