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Collection of Series for Π ∑ ∑ ∑ Collection of series for $\pi $ Page 1 of 9 Collection of series for πππ (Click here for a Postscript version of this page.) 1 Introduction There are a great many numbers of series involving the constant π, we provide a selection. The great Swiss mathematician Leonhard Euler (1707-1783) discovered many of those. 2 Around Leibniz-Gregory-Madhava series π 1 1 1 = 1− + − +... (Leibniz −Gregory −Madhava) 4 3 5 7 ∞ 2 k 1 1 π = ( −1) 1+ +...+ (Knopp) 16 ∑ k + 1 3 2k + 1 k = 0 π 3 1 1 1 = + − + −... (Nilakantha) 4 4 2.3.4 4.5.6 6.7.8 π 1 1.2 1.2.3 = 1+ + + +... (Euler) 2 3 3.5 3.5.7 π 1.2 1.2.3 1.2.3.4 = + + +... 2 1.3 1.3.5 1.3.5.7 ∞ 3k − 1 π = ζ(k + 1) (Flajolet −Vardi) ∑ k 4 k = 1 ∞ π 1 = ∑ arctan 2 (Knopp) 4 k + k + 1 k = 1 ∞ 1 1 π = tan (Euler) ∑ k + 1 k + 1 π 2 2 k = 1 π√2 1 1 1 1 1 = 1+ − − + + −... 4 3 5 7 9 11 π√3 1 1 1 1 1 = 1− + − + − +... 6 5 7 11 13 17 π√3 1 1 1 1 1 = 1− + − + − +... 9 2 4 5 7 8 ∞ π√3 (−1) k = (Sharp) 6 ∑ k 3 (2 k + 1) k = 0 3 Euler's series http://numbers.computation.free.fr/Constants/Pi/piSeries.html 03/12/2006 Collection of series for $\pi $ Page 2 of 9 It was a great problem to find the limit of the series 1 1 1 1+ + +...+ 2 +..., 4 9 k some of the greatest mathematicians of the seventeenth century failed to find this limit. After trying unsuccessfully to solve it, Jakob Bernoulli challenged mathematicians with this problem. It was Euler, in 1735, who found the value of the series and most of the following result are also due to him (see [ 4]). 3.1 All integers ∞ π2 1 1 1 = ∑ 2 =1+ 2 + 2 +... 6 k 2 3 k = 1 ∞ π4 1 1 1 = ∑ 4 =1+ 4 + 4 +... 90 k 2 3 k = 1 ∞ π6 1 1 1 = ∑ 6 =1+ 6 + 6 +... 945 k 2 3 k = 1 ∞ p 2p 4 | B 2p| π 1 = ∑ 2p = ζ (2 p) 2(2 p)! k k = 1 3.2 Odd integers ∞ π2 1 1 1 = ∑ 2 =1+ 2 + 2 +... 8 (2 k + 1) 3 5 k = 0 ∞ π4 1 1 1 = ∑ 4 =1+ 4 + 4 +... 96 (2 k + 1) 3 5 k = 0 ∞ π6 1 1 1 = ∑ 6 =1+ 6 + 6 +... 960 (2 k + 1) 3 5 k = 0 ∞ p 2p (4 − 1)| B 2p| π 1 = ∑ 2p 2(2 p)! (2 k + 1) k = 0 3.3 All integers alternating http://numbers.computation.free.fr/Constants/Pi/piSeries.html 03/12/2006 Collection of series for $\pi $ Page 3 of 9 ∞ π2 (−1) k + 1 1 1 = =1 − + −... ∑ 2 2 2 12 k 2 3 k = 1 ∞ 7π4 (−1) k + 1 1 1 = =1 − + −... ∑ 4 4 4 720 k 2 3 k = 1 ∞ 31 π6 (−1) k + 1 1 1 = =1 − + −... ∑ 6 6 6 30240 k 2 3 k = 1 ∞ (4 p − 2)| B | π2p ( −1) k + 1 2p = ∑ 2p 2(2 p)! k k = 1 3.4 Odd integers alternating ∞ π3 (−1) k 1 1 = =1 − + −... ∑ 3 3 3 32 (2 k + 1) 3 5 k = 0 ∞ 5π5 (−1) k 1 1 = =1 − + −... ∑ 5 5 5 1536 (2 k + 1) 3 5 k = 0 ∞ 61 π7 (−1) k 1 1 = =1 − + −... ∑ 7 7 7 184320 (2 k + 1) 3 5 k = 0 ∞ 2p + 1 | E | π (−1) k 2p = p + 1 ∑ (2 k + 1) 2p + 1 4 (2 p)! k = 0 Bn and E n are respectively Bernoulli's numbers and Euler's numbers. 1 1 1 1 1 5 B = 1, B = − , B = , B = − , B = , B = − , B = ,... 0 1 2 2 6 4 30 6 42 8 30 10 66 E0 = 1, E 2= −1, E 4=5, E 6= −61, E 8=1385, E 10 = −50251,... 3.5 With prime numbers In the following series, only the denominators with an odd number of prime factors are taken in account. For example 10=2×5 is omitted because it has two prime factors. http://numbers.computation.free.fr/Constants/Pi/piSeries.html 03/12/2006 Collection of series for $\pi $ Page 4 of 9 2 π 1 1 1 1 1 1 = 2 + 2 + 2 + 2 + 2 + 2 +... 20 2 3 5 7 8 11 4 π 1 1 1 1 1 1 = 4 + 4 + 4 + 4 + 4 + 4 +... 1260 2 3 5 7 8 11 4π6 1 1 1 1 1 1 = 6 + 6 + 6 + 6 + 6 + 6 +... 225225 2 3 5 7 8 11 ζ2(2 p) − ζ (4 p) 1 1 1 1 1 1 = 2p + 2p + 2p + 2p + 2p + 2p +... 2ζ(2 p) 2 3 5 7 8 11 If this time the prime factors are also supposed to be different: 9 1 1 1 1 1 1 2 = 2 + 2 + 2 + 2 + 2 + 2 +... 2π 2 3 5 7 11 13 15 1 1 1 1 1 1 4 = 4 + 4 + 4 + 4 + 4 + 4 +... 2π 2 3 5 7 11 13 11340 1 1 1 1 1 1 6 = 6 + 6 + 6 + 6 + 6 + 6 +... 691 π 2 3 5 7 11 13 2 ζ (2 p) − ζ (4 p) 1 1 1 1 1 1 = 2p + 2p + 2p + 2p + 2p + 2p +... 2ζ(2 p)ζ(4 p) 2 3 5 7 11 13 4 Machin's formulae By mean of the function 1 (−1) k L( p) = arctan = p ∑ 2k + 1 (2 k + 1) p k ≥ 0 numerous more or less efficient formulae to express π are available ( [ 1], [ 5], [ 7], [ 9]). Observe that the Leibniz-Gregory-Madhava series may be written as π/4 = L(1) and Sharp's series is just π/6 = L( √3). 4.1 Two terms formulae π = 2L( √2) + L(2 √2) (Wetherfield) 2 π = L(2) + L(3) (Hutton) 4 π = 2L(3) + L(7) (Hutton) 4 π = 4L(5) − L(239) (Machin) 4 π = 2L(3 √3) + L(4 √3) 6 http://numbers.computation.free.fr/Constants/Pi/piSeries.html 03/12/2006 Collection of series for $\pi $ Page 5 of 9 π = 5L(7) + 2L(79/3) (Euler) 4 π = 5L(278/29) + 7L(79/3) 4 4.2 Three terms and more formulae π = L(2) + L(5) + L(8) (Strassnitzky) 4 π = 4L(5) − L(70) + L(99) (Euler) 4 π = 5L(7) + 4L(53) + 2L(4443) 4 π = 6L(8) + 2L(57) + L(239) (Störmer) 4 π = 8L(10) − L(239) − 4L(515) (Klingenstierna) 4 π = 12L(18) + 8L(57) − 5L(239) (Gauss) 4 π = 22L(38) + 17L(601/7) + 10L(8149/7) (Sebah) 4 π = 44L(57) + 7L(239) − 12L(682) + 24L(12943) (Störmer) 4 π = 88L(172) + 51L(239) + 32L(682) + 44L(5357) + 68L(12943) (Störmer) 4 For example, more than 100 three terms formulae are known and are easy to generate by mean of dedicated algorithms. 5 BBP series In 1995, Bailey, Borwein and Plouffe ( BBP ) found a new kind of formula which allows to compute directly the d-th digit of π in basis 2 (see [ 2]) ∞ 4 2 1 1 1 π = ∑ − − − k . 8k + 1 8k + 4 8k + 5 8k + 6 16 k = 0 Other such formulae are available: ∞ 2 2 1 ( −1) k π = + + , ∑ 4k + 1 4k + 2 4k + 3 k 4 k = 0 ∞ http://numbers.computation.free.fr/Constants/Pi/piSeries.html 03/12/2006 Collection of series for $\pi $ Page 6 of 9 π = 2 1 1 1 1 1 1 ∑ + + − − − , 8k+1 4k+1 8k+3 16k+10 16k+12 32k+28 16 k k = 0 ∞ 1 32 1 256 64 4 4 1 ( −1) k π = − − + − − − + , 64 ∑ 4k+1 4k+3 10k+1 10k+3 10k+5 10k+7 10k+9 k 1024 k = 0 ∞ 4 1 1 ( −1) k π√2 = + + , ∑ 6k+1 6k+3 6k+5 k 8 k = 0 ∞ 8 π2 16 24 8 6 1 1 = ∑ 2 − 2 − 2 − 2 + 2 k . 9 (6k+1) (6k+2) (6k+3) (6k+4) (6k+5) 64 k = 0 The series with 1024 k is efficient and due to F. Bellard (1997). 6 Ramanujan's series Most of those series and many others were found by the Indian prodigy Srinivasa Ramanujan (1887-1920) ([ 3], [ 8]). 3 3 3 2 1 1.3 1.3.5 = +9 +... 1−5 −13 π 2 2.4 2.4.6 2 2 2 2 4 1 1 1.3 1.3.5 = 1+ + + + +... (Forsyth) π 2 2.4 2.4.6 2.4.6.8 ∞ 1 (2k)! 3 (42k + 5) = π ∑ 6 12k + 4 (k!) 2 k = 0 ∞ 1 1 k (4k)! (23 + 260k) = ∑ (−1) 4 4k 2k π 72 (k!) 4 18 k = 0 ∞ 1 1 k (4k)! (1123 + 21460k) = ∑ (−1) 4 4k 2k π 3528 (k!) 4 882 k = 0 ∞ 1 2√2 (4k)! (1103 + 26390k) = ∑ 4 4k 4k π 9801 (k!) 4 99 k = 0 ∞ 1 k (6k)! (13591409 + 545140134k) = 12 ∑ (−1) 3 3k + 3/2 (Chudnovsky) π (3k)!(k!) 640320 k = 0 ∞ http://numbers.computation.free.fr/Constants/Pi/piSeries.html 03/12/2006 Collection of series for $\pi $ Page 7 of 9 1 = (6k)! (A + Bk) π 12 ∑ (−1) k (Borwein) (3k)!(k!) 3 C3k + 3/2 k = 0 In the last formula 61 A = 1657145277365+212175710912 √ , 61 B = 107578229802750+13773980892672 √ , 61 C = 5280(236674+30303 √ ), and each additional term in the series adds about 31 digits ..
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