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Collection of Series for Π Collection of series for π Pascal Sebah and Xavier Gourdon numbers.computation.free.fr/Constants/constants.html March 31, 2004 Abstract Selection of some of the numerous series expansion involving the fa- mous constant π. 1 Introduction There are a great many numbers of series involving the constant π,weprovide a selection. The celebrated 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 π2 − k ( 1) 1 ··· 1 = k 1+ + + k (Knopp) 16 k≥0 +1 3 2 +1 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 k − π 3 1ζ k = k ( + 1) (Flajolet-Vardi) k≥1 4 π 1 = arctan k2 k (Knopp) 4 k≥1 + +1 π 1 = arctan F2k+1 4 k≥1 1 1 1 π π = k+1 tan k+1 (Euler) k≥1 2 2 √ 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 = k k (Sharp) 6 k≥0 3 (2 +1) The Fn are Fibonacci numbers. F0 =0,F1 =1,F2 =1,F3 =2,F4 =3,F5 =5, ..., Fn+1 = Fn + Fn−1. 3 Euler’s series It was a great problem to find the limit of the series 1 1 1 1+ + + ···+ + ··· , 4 9 k2 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 ··· = k2 =1+ 2 + 2 + 6 k≥1 2 3 π4 1 1 1 ··· = k4 =1+ 4 + 4 + 90 k≥1 2 3 π6 1 1 1 ··· = k6 =1+ 6 + 6 + 945 k≥1 2 3 p |B | π2p 4 2p 1 ζ p p = k2p = (2 ) 2(2 )! k≥1 2 3.2 Odd integers π2 1 1 1 ··· = k 2 =1+ 2 + 2 + 8 k≥0 (2 +1) 3 5 π4 1 1 1 ··· = k 4 =1+ 4 + 4 + 96 k≥0 (2 +1) 3 5 π6 1 1 1 ··· = k 6 =1+ 6 + 6 + 960 k≥0 (2 +1) 3 5 p 2p (4 − 1) |B2p| π 1 p = k 2p 2(2 )! k≥0 (2 +1) 3.3 All integers alternating π2 − k+1 ( 1) − 1 1 −··· = k2 =1 2 + 2 12 k≥1 2 3 π4 − k+1 7 ( 1) − 1 1 −··· = k4 =1 4 + 4 720 k≥1 2 3 π6 − k+1 31 ( 1) − 1 1 −··· = k6 =1 6 + 6 30240 k≥1 2 3 p 2p k+1 (4 − 2) |B2p| π (−1) p = k2p 2(2 )! k≥1 3.4 Odd integers alternating π3 − k ( 1) − 1 1 −··· = k 3 =1 3 + 3 32 k≥0 (2 +1) 3 5 π5 − k 5 ( 1) − 1 1 −··· = k 5 =1 5 + 5 1536 k≥0 (2 +1) 3 5 π7 − k 61 ( 1) − 1 1 −··· = k 7 =1 7 + 7 184320 k≥0 (2 +1) 3 5 2p+1 k |E2p| π (−1) p+1 p = k 2p+1 4 (2 )! k≥0 (2 +1) Bn and En are respectively Bernoulli’s numbers and Euler’s numbers: 3 1 1 1 1 1 5 B0 =1,B1 = − ,B2 = ,B4 = − ,B6 = ,B8 = − ,B10 = , ..., 2 6 30 42 30 66 E0 =1,E2 = −1,E4 =5,E6 = −61,E8 = 1385,E10 = −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 the number 10 = 2 × 5 is omitted because it has two prime factors. π2 1 1 1 1 1 1 = + + + + + + ··· 20 22 32 52 72 82 112 π4 1 1 1 1 1 1 = + + + + + + ··· 1260 24 34 54 74 84 114 4π6 1 1 1 1 1 1 = + + + + + + ··· 225225 26 36 56 76 86 116 ζ2(2p) − ζ(4p) 1 1 1 1 1 1 = + + + + + + ··· 2ζ(2p) 22p 32p 52p 72p 82p 112p If this time the prime factors are also supposed to be different: 9 1 1 1 1 1 1 = + + + + + + ··· 2π2 22 32 52 72 112 132 15 1 1 1 1 1 1 = + + + + + + ··· 2π4 24 34 54 74 114 134 11340 1 1 1 1 1 1 = + + + + + + ··· 691π6 26 36 56 76 116 136 ζ2(2p) − ζ(4p) 1 1 1 1 1 1 = + + + + + + ··· 2ζ(2p)ζ(4p) 22p 32p 52p 72p 112p 132p 4 Machin’s formulae By mean of the function − k p 1 ( 1) L( ) = arctan p = k p2k+1 k≥0 (2 +1) numerous more or less efficient formulae to express π are available. (Consult [1], [5], [7], [9].) π Observe that the Leibniz-Gregory-Madhava√series may be written as 4 = π L(1) and Sharp’s series is equivalent to 6 =L( 3). 4 4.1 Two terms formulae π √ √ =2L(2) + L(2 2) (Wetherfield) 2 π = L(2) + L(3) (Hutton) 4 π = 2 L(3) + L(7) (Hutton) 4 π = 4 L(5) − L(239) (Machin) 4 π √ √ =2L(33) + L(4 3) 6 π 79 = 5 L(7) + 2 L (Euler) 4 3 π 278 79 =5L +7L 4 29 3 4.2 Three terms and more formulae π = L(2) + L(5) + L(8) (Strassnitzky) 4 π = 4 L(5) − L(70) + L(99) (Euler) 4 π = 5 L(7) + 4 L(53) + 2 L(4443) 4 π = 6 L(8) + 2 L(57) + L(239) (St¨ormer) 4 π = 8 L(10) − L(239) − 4 L(515) (Klingenstierna) 4 π = 12 L(18) + 8 L(57) − 5 L(239) (Gauss) 4 π = 16 L(21) + 3 L(239) + 4 L(1042/3) 4 π = 22 L(28) + 2 L(443) − 5 L(1393) − 10 L(11018) 4 π = 22 L(38) + 17 L(601/7) + 10 L(8149/7) (Sebah) 4 π = 44 L(57) + 7 L(239) − 12 L(682) + 24 L(12943) (St¨ormer) 4 π = 88 L(172) + 51 L(239) + 32 L(682) + 44 L(5357) + 68 L(12943) (St¨ormer) 4 For example, more than 100 three terms formulae are known and are easy to generate by mean of dedicated algorithms. 5 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].) π 4 − 2 − 1 − 1 1 . = k k k k k k≥0 8 +1 8 +4 8 +5 8 +6 16 Other such formulae are available. − k π 2 2 1 ( 1) = k + k + k k k≥0 4 +1 4 +2 4 +3 4 π 2 1 1 − 1 − 1 − 1 1 = k + k + k k k k k k≥0 8 +1 4 +1 8 +3 16 +10 16 +12 32 +28 16 − k π 1 − 32 − 1 256 − 64 − 4 − 4 1 ( 1) = k k + k k k k + k k 64 k≥0 4 +1 4 +3 10 +1 10 +3 10 +5 10 +7 10 +9 1024 √ − k π 4 1 1 ( 1) 2= k + k + k k k≥0 6 +1 6 +3 6 +5 8 π2 8 16 − 24 − 8 − 6 1 1 = k 2 k 2 k 2 k 2 + k 2 k 9 k≥0 (6 +1) (6 +2) (6 +3) (6 +4) (6 +5) 64 The series with 1024k is efficient and due to Fabrice 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 =1− 5 +9 − 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 3 1 2k (42k +5) π = k 12k+4 k≥0 2 k k 1 1 − k (4 )! (23 + 260 ) π = ( 1) k 4 4k 2k 72 k≥0 ( !) 4 18 k k 1 1 − k (4 )! (1123 + 21460 ) π = ( 1) k 4 4k 2k 3528 k≥0 ( !) 4 882 6 √ 1 2 2 (4k)! (1103 + 26390k) π = k 4 4k 4k 9801 k≥0 ( !) 4 99 k k 1 − k (6 )! (13591409 + 545140134 ) π =12 ( 1) k k 3 3k+3/2 (Chudnovsky) k≥0 (3 )!( !) 640320 1 (6k)! (A + Bk) =12 (−1)k (Borwein) π k k 3 C3k+3/2 k≥0 (3 )!( !) In the last formula √ A , = 1657145277365 + 212175710912 61 √ B , = 107578229802750 + 13773980892672√ 61 C = 5280(236674 + 30303 61), and each additional term in the series adds about 31 digits! 7 Other series π − 3 (−1)k+1 = k2 − 6 k≥1 36 1 − k+1 π − ( 1) 3= k k k k≥1 ( + 1)(2 +1) π3 +8π − 56 (−1)k+1 = k k k 3 8 k≥1 ( + 1)(2 +1) π (−1)(k−1)/2 = k k4 (Glaisher) 16 k odd ( +4) π − 1 =1 16 k 2 k 2 k 2 (Lucas) 4 k≥0 (4 +1) (4 +3) (4 +5) − π2 1 10 = k3 k 3 k≥1 ( +1) π2 − 8 1 = k2 − 2 (Euler) 16 k≥1 (4 1) 32 − 3π2 1 = k2 − 3 (Euler) 64 k≥1 (4 1) π4 +30π2 − 384 1 = k2 − 4 (Euler) 768 k≥1 (4 1) 7 √ π k 2 2 3 ! 1 1 1 ··· = k =1+ + + + 9 k≥0 (2 + 1)! 6 30 140 √ π k 2 2 3 4 ! 1 1 1 1 ··· + = k =1+ + + + + 27 3 k≥0 (2 )! 2 6 20 70 √ π 3 k!2 = k k 9 k≥1 (2 )! π2 k!2 = k k2 (Euler) 18 k≥1 (2 )! 17π4 k!2 = k k4 (Comtet) 3240 k≥1 (2 )! π (2k)! = k kk 2 3 k≥0 (2 + 1)16 ! k 2k k π ! 2 +3 = k k≥1 (2 )! k − k k π (25 3) !(2 )! = k−1 k (Gosper 1974) k≥0 2 (3 )! k − 8 (4 +3)C4 1 π2 = 8k+4 k (Victor) k≥0 2 Cn are Catalan’s numbers: (2n)! C0 =1,C1 =1,C2 =2,C3 =5,C4 =14,C5 =42, ..., Cn = .
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