Pi = 4*Sum_{k>=0} (-1)^k/(2k+1) [Madhava-Gregory-Leibniz, 1450-1671]. -
N. J. A. Sloane, Feb 27 2013
2/Pi = (sqrt(2)/2) * (sqrt(2 + sqrt(2))/2) * (sqrt(2 + sqrt(2 + sqrt(2)))/2) * ... [Viete, 1593]
2/Pi = Product_{k>=1} (4*k^2-1)/(4*k^2). [Wallis, 1655]
Pi = 3*sqrt(3)/4 + 24*(1/12 - Sum_{n>=2} (2*n-2)!/((n-1)!^2*(2*n-3)*(2*n+1)*2^(4*n-2))). [Newton, 1666]
Pi/4 = 4*arctan(1/5) - arctan(1/239). [Machin, 1706]
Pi^2/6 = 3*Sum_{n>=1} 1/(n^2*binomial(2*n,n)). [Euler, 1748]
1/Pi = (2*sqrt(2)/9801) * Sum_{n>=0} (4*n)!*(1103+26390*n)/((n!)^4*396^(4*n)). [Ramanujan, 1914]
1/Pi = 12*Sum_{n>=0} (-1)^n*(6*n)!*(13591409 + 545140134*n)/((3*n)!*(n!)^3*(640320^3)^(n+1/2)). [David and Gregory Chudnovsky, 1989]
Pi = Sum_{n>=0} (1/16^n) * (4/(8*n+1) - 2/(8*n+4) - 1/(8*n+5) - 1/(8*n+6)). [Bailey-Borwein-Plouffe, 1989] (End)
Pi - 2 = 1/1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - 1/28 - 1/36 + 1/45 + ... [Jonas Castillo Toloza, 2007], that is, Pi - 2 = Sum_{n>=1} (1/((-1)^floor((n-1)/2)*(n^2+n)/2)). -
José de Jesús Camacho Medina, Jan 20 2014
Pi = 3 * Product_{t=img(r),r=(1/2+i*t) root of zeta function} (9+4*t^2)/(1+4*t^2) <=> RH is true. -
Dimitris Valianatos, May 05 2016
Pi = Sum_{k>=1} (3^k - 1)*zeta(k+1)/4^k.
Pi = 2*Product_{k>=2} sec(Pi/2^k).
Pi = 2*Integral_{x>=0} sin(x)/x dx. (End)
Pi = 2^{k + 1}*arctan(sqrt(2 - a_{k - 1})/a_k) at k >= 2, where a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2). -
Sanjar Abrarov, Feb 07 2017
Pi = lim_{n->infinity} 2/n * Sum_{m=1,n} ( sqrt( (n+1)^2 - m^2 ) - sqrt( n^2 - m^2 ) ). -
Dimitri Papadopoulos, May 31 2019
Pi = Sum_{n >= 0} 2^(n+1)/( binomial(2*n,n)*(2*n + 1) ) - Euler.
More generally, Pi = (4^x)*x!/(2*x)! * Sum_{n >= 0} 2^(n+1)*(n+x)!*(n+2*x)!/(2*n+2*x+1)! = 2*4^x*x!^2/(2*x+1)! * hypergeom([2*x+1,1], [x+3/2], 1/2), valid for complex x not in {-1,-3/2,-2,-5/2,...}. Here, x! is shorthand notation for the function Gamma(x+1). This identity may be proved using Gauss's second summation theorem.
Setting x = 3/4 and x = -1/4 (resp. x = 1/4 and x = -3/4) in the above identity leads to series representations for the constant
A085565 (resp.
A076390). (End)
Equals 2 + Integral_{x=0..1} arccos(x)^2 dx.
Equals Integral_{x=0..oo} log(1 + 1/x^2) dx.
Equals Integral_{x=0..oo} log(1 + x^2)/x^2 dx.
Equals Integral_{x=-oo..oo} exp(x/2)/(exp(x) + 1) dx. (End)
Pi = 32*Sum_{n >= 1} (-1)^n*n^2/((4*n^2 - 1)*(4*n^2 - 9))= 384*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)*(4*n^2 - 9)*(4*n^2 - 25)).
More generally, it appears that for k = 1,2,3,..., Pi = 16*(2*k)!*Sum_{n >= 1} (-1)^(n+k+1)*n^2/((4*n^2 - 1)* ... *(4*n^2 - (2*k+1)^2)).
Pi = 32*Sum_{n >= 1} (-1)^(n+1)*n^2/(4*n^2 - 1)^2 = 768*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)^2*(4*n^2 - 9)^2).
More generally, it appears that for k = 0,1,2,..., Pi = 16*Catalan(k)*(2*k)!*(2*k+2)!*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)^2* ... *(4*n^2 - (2*k+1)^2)^2).
Pi = (2^8)*Sum_{n >= 1} (-1)^(n+1)*n^2/(4*n^2 - 1)^4 = (2^17)*(3^5)*Sum_{n >= 2} (-1)^n*n^2*(n^2 - 1)/((4*n^2 - 1)^4*(4*n^2 - 9)^4) = (2^27)*(3^5)*(5^5)* Sum_{n >= 3} (-1)^(n+1)*n^2*(n^2 - 1)*(n^2 - 4)/((4*n^2 - 1)^4*(4*n^2 - 9)^4*(4*n^2 - 25)^4). (End)
Pi = 4/phi + Sum_{n >= 0} (1/phi^(12*n)) * ( 8/((12*n+3)*phi^3) + 4/((12*n+5)*phi^5) - 4/((12*n+7)*phi^7) - 8/((12*n+9)*phi^9) - 4/((12*n+11)*phi^11) + 4/((12*n+13)*phi^13) ) where phi = (1+sqrt(5))/2. -
Chittaranjan Pardeshi, May 16 2022
Pi = 48*Sum_{n >= 0} (-1)^n/((6*n + 1)*(6*n + 3)*(6*n + 5)).
More generally, it appears that for k >= 0 we have Pi = A(k) + B(k)*Sum_{n >= 0} (-1)^n/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 5)), where A(k) is a rational approximation to Pi and B(k) = (3 * 2^(3*k+3) * (3*k + 2)!) / (2^(3*k+1) - (-1)^k). The first few values of A(k) for k >= 0 are [0, 256/85, 65536/20955, 821559296/261636375, 6308233216/2008080987, 908209489444864/289093830828075, ...].
Pi = 16/5 - (288/5)*Sum_{n >= 0} (-1)^n * (6*n + 1)/((6*n + 1)*(6*n + 3)*...*(6*n + 9)).
More generally, it appears that for k >= 0 we have Pi = C(k) + D(k)*Sum_{n >= 0} (-1)^n* (6*n + 1)/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 3)), where C(k) and D(k) are rational numbers. The case k = 0 is the Madhava-Gregory-Leibniz series for Pi.
Pi = 168/53 + (288/53)*Sum_{n >= 0} (-1)^n * (42*n^2 + 25*n)/((6*n + 1)*(6*n + 3)*(6*n + 5)*(6*n + 7)).
More generally, it appears that for k >= 1 we have Pi = E(k) + F(k)*Sum_{n >= 0} (-1)^n * (6*(6*k + 1)*n^2 + (24*k + 1)*n)/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 1)), where E(k) and F(k) are rational numbers. (End)
The series representation Pi = 4 * Sum_{k >= 0} 1/(4*k + 1) - 1/(4*k + 3) given above by
Alexander R. Povolotsky, Dec 25 2008, is the case n = 0 of the more general result (obtained by the WZ method): for n >= 0, there holds
Pi = Sum_{j = 0.. n-1} 2^(j+1)/((2*j + 1)*binomial(2*j,j)) + 8*(n+1)!*Sum_{k >= 0} 1/((4*k + 1)*(4*k + 3)*...*(4*k + 2*n + 3)).
Letting n -> oo gives the rapidly converging series Pi = Sum_{j >= 0} 2^(j+1)/((2*j + 1)*binomial(2*j,j)) due to Euler.
More generally, it appears that for n >= 1, Pi = 1/(2*n-1)!!^2 * Sum_{j >= 0} (Product_{i = 0..2*n-1} j - i) * 2^(j+1)/((2*j + 1)*binomial(2*j,j)).
For any integer n, Pi = (-1)^n * 4 * Sum_{k >= 0} 1/(4*k + 1 + 2*n) - 1/(4*k + 3 - 2*n). (End)
Equals Integral_{x=0..2} sqrt(8 - x^2) dx - 2 (see Ambrisi and Rizzi). -
Stefano Spezia, Jul 21 2024
Equals 3 + 4*Sum_{k>0} (-1)^(k+1)/(4*k*(1+k)*(1+2*k)) (see Wells at p. 53). -
Stefano Spezia, Aug 31 2024
Equals 4*Integral_{x=0..1} sqrt(1 - x^2) dx = lim_{n->oo} (4/n^2)*Sum_{k=0..n} sqrt(n^2 - k^2) (see Finch). -
Stefano Spezia, Oct 19 2024
