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A quick perusal of the wikipedia page for $\pi$ yields a large collection of known series for $\pi$. In particular, these series are hypergeometric in nature, and have large (but finite) radius of convergence.

My first question is, what is the best known series for $\pi$? Namely, a series of the form above with the largest radius of convergence.

My second question is one asked by Herbert Wilf (http://www.math.upenn.edu/~wilf/website/UnsolvedProblems.pdf), which is to ask whether there exists a sequence $(a_n)$ with $a_{n+1}/a_n$ a rational functional of $n$ for all $n$, and the function $f(z) = \displaystyle \sum_{n=0}^\infty a_n z^n$ is an entire function, and $f(1) = \pi$. Presumably, such a function is not known to exist yet. Can anyone give any recent works that advances our understanding on this problem, or give some insight as to why such a function is so difficult to find?

Thanks!

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    $\begingroup$ I'm not sure I get your question. Why look at the radius of convergence? $\endgroup$
    – Thierry Zell
    Feb 22, 2011 at 22:00
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    $\begingroup$ @Thierry: a large ROC means that $a_n\to0$ rapidly, so that the partial sums are good approximations. A more fundamental phrasing of the question would be to ask for the function $\alpha(n)$ that can be computed with the fewest flops (asymptotically) and which satisfies $|\pi-\alpha(n)|<1/n$. $\endgroup$
    – Kevin O'Bryant
    Feb 22, 2011 at 22:11
  • $\begingroup$ But if $\sum a_nz^n$ has radius of convergence $r$, then $\sum b_nz^n$, where $b_n=a_n/2^n$, has radius of convergence $2r$, and doesn't do a better job of getting $\pi$. Are we assuming that $\pi$ is necessarily $f(1)$? $\endgroup$
    – Gerry Myerson
    Feb 22, 2011 at 22:15
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    $\begingroup$ Depends what you mean by best? If you want to find a specific digit, then something like the BBP formula should suit your needs. Otherwise the Chudnovsky modification of Ramanujan's formula is very fast. $\endgroup$
    – Alex R.
    Feb 22, 2011 at 22:30
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    $\begingroup$ The question should be rephrased. Instead of "My first question is, what is the best known series for $\pi$? Namely, a series of the form above with the largest radius of convergence", I suggest: "Among all power series $f(x)=\sum_{n=0}^\infty a_n x^n$ with $f(1)=\pi$, and with the extra property that $a_{n+1}/a_n$ is a rational expression of $n$, which one has the largest radius of convergence?" $\endgroup$
    – André Henriques
    Feb 22, 2011 at 23:19

3 Answers 3

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I like Andre Henriques' rephrasing. The Borwein, Bailey, Plouffe series, with $$a_n={1\over16^n}\left({4\over8n+1}-{2\over8n+4}-{1\over8n+5}-{1\over8n+6}\right)$$ would have radius of convergence $r=16$. Bellard gives a more complicated one with $r=1024$. Pschill has one with 21 terms and $r=2^{30}$. If you'll accept $f(1)=1/\pi$, D and G Chudnovsky give $$a_n={12\over\sqrt{640320^3}}(-1)^n{(6n)!\over(n!)^3(3n)!}{13591409+54514013n\over(640320^3)^n}$$ All of these are taken from Chapter 16 of Arndt and Haenel, $\pi$ Unleashed, which gives full bibliographic citations.

EDIT: See also http://mathworld.wolfram.com/PiFormulas.html in particular formulas 93-96 where each term gives another 50 digits of $1/\pi$ (which I guess corresponds to $r$ roughly $10^{50}$). Somehow, the series for $1/\pi$ seem to do better than those for $\pi$. I know there are people who would like us to abandon $\pi$ in favor of $2\pi$, but maybe we should really be expressing things in terms of $1/(2\pi)$.

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In Almkvist-Krattenthaler-Petersson:"Some new formulas for Pi" arXiv 2003?, Exp. Math 12(2003) 441-456 it is shown that there exist formulas for Pi where each term gives N new digits for any given N. Gert Almkvist

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    $\begingroup$ @Gert, welcome to MO! $\endgroup$
    – Gerry Myerson
    Apr 2, 2011 at 22:45
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Paper https://arxiv.org/abs/1706.08835 describes and shows how we can get any N correct digits of pi just per term increment. This Machin-like formula for pi is rapid in convergence and without irrational numbers.

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  • $\begingroup$ I couldn't find that result in the paper, but anyway such a result was already published in 2003, according to the answer posted by Gert Almkvist earlier on this question. $\endgroup$
    – Gerry Myerson
    Sep 25, 2017 at 8:07
  • $\begingroup$ @ Gerry Myerson. Let me clarify. At different k in the eq. (24) we can get different number N of corrects digits of pi. This tendency is shown in Fig. 1. Therefore at larger k the eq. (24) becomes a rapid series for pi (N correct digits of pi per term increases with increasing k, see also Mathematica code in the paper as an example). $\endgroup$
    – Philip Thomas
    Sep 25, 2017 at 13:56

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