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

1,1

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

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

Cf. A002388, A013661, A019669.

Sequence in context: A142473 A132426 A072646 * A257578 A272665 A114431

Adjacent sequences: A091473 A091474 A091475 * A091477 A091478 A091479

Keyword

nonn,cons

Author

Eric W. Weisstein, Jan 13 2004

Status

approved