A091476 Decimal expansion of Pi^2/4.
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
Examples
2.46740110027233965470862274996903778...
Links
- Muniru A Asiru, Table of n, a(n) for n = 1..2000
- 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.
- T. Piezas III, Golden ratio and nested radicals
- Johan Wästlund, Summing inverse squares by euclidean geometry
- 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.
- Index entries for transcendental numbers
Programs
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Maple
evalf(Pi^2/4, 120); # Muniru A Asiru, Sep 18 2018
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Mathematica
First[RealDigits[Pi^2/4,10,100]] (* Paolo Xausa, Oct 31 2023 *)
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PARI
Pi^2/4 \\ Charles R Greathouse IV, Mar 02 2018
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PARI
2*sumpos(n=1,(2*n-1)^-2) \\ Charles R Greathouse IV, Mar 02 2018
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 ... + (-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
From A.H.M. Smeets, Apr 10 2025: (Start)
Let X(p,q) be the p-th smallest zero of the Laguerre polynomial of order q.
Equals lim_{k -> oo} X(k,k^2).
Equals lim_{q -> oo} X(1,q)*q.
Equals lim_{k -> oo} X(k,k^4)*sqrt(k).
Equals lim_{k -> oo} X(k^3,k^4)/sqrt(k).
More general, let P = log_q(p^2/q), then, for any p, 0 < p <= q, equals lim_{q -> oo} X(p,q)/q^P. (End)
Equals Integral_{x=-1..1} -log(abs(x))/(1 - x^2) dx. - Kritsada Moomuang, May 28 2025