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%I #85 Jan 28 2024 18:43:24
%S 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,
%T 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,
%U 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
%N Decimal expansion of Pi^2/4.
%H Muniru A Asiru, <a href="/A091476/b091476.txt">Table of n, a(n) for n = 1..2000</a>
%H Ben Hambrecht and Grant Sanderson, <a href="https://www.youtube.com/watch?v=d-o3eB9sfls">The stunning geometry behind this surprising equation</a>, 3Blue1Brown video (2018).
%H Josef Hofbauer, <a href="https://www.jstor.org/stable/2695334">A simple proof of 1 + 1/2^2 + 1/3^2 + ... = Pi^2/6 and related identities</a>, The American Mathematical Monthly 109:2 (2002), pp. 196-200.
%H Michael Penn, <a href="https://www.youtube.com/watch?v=CGcY6zzBBm0">Neat ways to solve complicated limits.</a>, YouTube video, 2023.
%H T. Piezas III, <a href="https://sites.google.com/view/tpiezas/0014-article-4-golden-ratio-and-nested-radicals">Golden ratio and nested radicals</a>
%H Johan Wästlund, <a href="http://www.math.chalmers.se/~wastlund/Cosmic.pdf">Summing inverse squares by euclidean geometry</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DefiniteIntegral.html">Definite Integral</a>
%H H. Wilf, <a href="http://www.emis.de/journals/DMTCS/volumes/abstracts/dm030406.abs.html">Accelerated series for universal constants, by the WZ method</a>, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F Equals Integral_{x=0..Pi} x*sin(x)/(1+cos(x)^2) dx.
%F Equals Integral_{x=0..1} log((1+x)/(1-x))/x dx. - _Jean-François Alcover_, May 13 2013
%F 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
%F Equals A003881 * A000796. - _R. J. Mathar_, Oct 09 2015
%F Equals ... + (-5)^-2 + (-3)^-2 + (-1)^-2 + 1^-2 + 3^-2 + 5^-2 + .... - _Charles R Greathouse IV_, Mar 02 2018
%F From _A.H.M. Smeets_, Sep 18 2018: (Start)
%F Equals A102753/2.
%F Equals 2*Sum_{k > 0} 1/(2*k - 1)^2. (End)
%F 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
%F Equals Integral_{x=0..oo} log(x)/(x^2 - 1) dx. - _Amiram Eldar_, Aug 12 2020
%F 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
%F From _Peter Bala_, Nov 16 2023: (Start)
%F 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.
%F 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
%F 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)
%e 2.46740110027233965470862274996903778...
%p evalf(Pi^2/4, 120); # _Muniru A Asiru_, Sep 18 2018
%t First[RealDigits[Pi^2/4,10,100]] (* _Paolo Xausa_, Oct 31 2023 *)
%o (PARI) Pi^2/4 \\ _Charles R Greathouse IV_, Mar 02 2018
%o (PARI) 2*sumpos(n=1,(2*n-1)^-2) \\ _Charles R Greathouse IV_, Mar 02 2018
%Y Cf. A002388, A013661, A019669.
%K nonn,cons
%O 1,1
%A _Eric W. Weisstein_, Jan 13 2004
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