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A091476
Decimal expansion of Pi^2/4.
20
2, 4, 6, 7, 4, 0, 1, 1, 0, 0, 2, 7, 2, 3, 3, 9, 6, 5, 4, 7, 0, 8, 6, 2, 2, 7, 4, 9, 9, 6, 9, 0, 3, 7, 7, 8, 3, 8, 2, 8, 4, 2, 4, 8, 5, 1, 8, 1, 0, 1, 9, 7, 6, 5, 6, 6, 0, 3, 3, 3, 7, 3, 4, 4, 0, 5, 5, 0, 1, 1, 2, 0, 5, 6, 0, 4, 8, 0, 1, 3, 1, 0, 7, 5, 0, 4, 4, 3, 3, 5, 0, 9, 2, 9, 6, 3, 8, 0, 5, 7, 9, 5
OFFSET
1,1
LINKS
Ben Hambrecht and Grant Sanderson, The stunning geometry behind this surprising equation, 3Blue1Brown video (2018).
Josef Hofbauer, A simple proof of 1 + 1/2^2 + 1/3^2 + ... = Pi^2/6 and related identities, The American Mathematical Monthly 109:2 (2002), pp. 196-200.
Michael Penn, Neat ways to solve complicated limits., YouTube video, 2023.
Eric Weisstein's World of Mathematics, Definite Integral
H. Wilf, Accelerated  series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.
FORMULA
Equals Integral_{x=0..Pi} x*sin(x)/(1+cos(x)^2) dx.
Equals Integral_{x=0..1} log((1+x)/(1-x))/x dx. - Jean-François Alcover, May 13 2013
Equals Integral_{x=0..oo} K_0(x)^2 dx, where K_0 is a modified Bessel function (see Gradstein-Ryshik 6.576.4). - R. J. Mathar, Oct 09 2015
Equals A003881 * A000796. - R. J. Mathar, Oct 09 2015
Equals ... + (-5)^-2 + (-3)^-2 + (-1)^-2 + 1^-2 + 3^-2 + 5^-2 + .... - Charles R Greathouse IV, Mar 02 2018
From A.H.M. Smeets, Sep 18 2018: (Start)
Equals A102753/2.
Equals 2*Sum_{k > 0} 1/(2*k - 1)^2. (End)
Pi^2/4 = Integral_{x = 0..oo} x/sinh(x) dx. More generally, Pi^2/4 = 2*(1 + 1/3^2 + ... + 1/(2*n-1)^2) + Integral_{x = 0..oo} exp(-2*n*x)*x/sinh(x). - Peter Bala, Nov 05 2019
Equals Integral_{x=0..oo} log(x)/(x^2 - 1) dx. - Amiram Eldar, Aug 12 2020
Equals Sum_{n >= 0} 2^(n+1)/((n+1)^2*binomial(2*n+1,n)). See my entry in A002544 dated Apr 18 2017. Cf. A253191. - Peter Bala, Jan 30 2023
From Peter Bala, Nov 16 2023: (Start)
Pi^2/4 = 16*Sum_{k >= 1} k^2/(4*k^2 - 1)^2 = (2*16^2)*Sum_{k >= 1} k^2/((4*k^2 - 1)*(4*k^2 - 9))^2.
The general result, which can be proved using the WZ method (see Wilf for examples of this method), is that for n >= 0 there holds
Pi^2/4 = 16^(n+1)*(2*n + 1)*(2*n)!^4/(4*n)! * Sum_{k >= 1} k^2/( (4*k^2 - 1)*(4*k^2 - 9)*...*(4*k^2 - (2*n+1)^2) )^2. (End)
Equals Re(Polylog(2, 2)). - Mohammed Yaseen, Jul 03 2024
EXAMPLE
2.46740110027233965470862274996903778...
MAPLE
evalf(Pi^2/4, 120); # Muniru A Asiru, Sep 18 2018
MATHEMATICA
First[RealDigits[Pi^2/4, 10, 100]] (* Paolo Xausa, Oct 31 2023 *)
PROG
(PARI) Pi^2/4 \\ Charles R Greathouse IV, Mar 02 2018
(PARI) 2*sumpos(n=1, (2*n-1)^-2) \\ Charles R Greathouse IV, Mar 02 2018
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Eric W. Weisstein, Jan 13 2004
STATUS
approved