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A181122 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).

%I A181122 #40 Mar 04 2024 01:17:36
%S A181122 8,8,8,3,1,3,5,7,2,6,5,1,7,8,8,6,3,8,0,4,0,7,5,5,2,2,7,0,2,0,3,7,9,3,
%T A181122 4,6,2,7,8,1,1,0,8,3,0,7,7,5,4,5,8,1,7,1,2,0,5,9,7,0,6,8,2,0,8,4,7,6,
%U A181122 9,9,0,6,9,6,4,0,4,2,3,8,0,4,1,5,8,1,9,7,3,6,7,1,9,2,4,2,0,4,5,9,7,0,7,6,6
%N A181122 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).
%H A181122 Gheorghe Coserea, <a href="/A181122/b181122.txt">Table of n, a(n) for n = 0..1000</a>
%H A181122 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.
%F A181122 Sum_{k>=0} (-1)^k/(5k+1) = Integral_{x=0..1}dx/(1+x^5) = (1/10)*sqrt(10-2*sqrt(5))*arctan((3/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*sqrt(10-2*sqrt(5))*arctan(-(1/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*sqrt(10-2*sqrt(5))*sqrt(5)*arctan(-(1/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*log(2)*sqrt(5) + (1/5)*log(2) - (1/20)*log(7-3*sqrt(5))*sqrt(5).
%F A181122 Equals Pi*sqrt(phi)/5^(5/4) + log(phi)/sqrt(5) + log(2)/5, where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Nov 01 2015
%F A181122 From _Peter Bala_, Feb 19 2024: (Start)
%F A181122 Equals (1/2)*Sum_{n >= 0} n!*(5/2)^n/(Product_{k = 0..n} 5*k + 1) = (1/2)*Sum_{n >= 0} n!*(5/2)^n/A008548(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(5*k + 1)).
%F A181122 Continued fraction: 1/(1 + 1^2/(5 + 6^2/(5 + 11^2/(5 + ... + (5*n + 1)^2/(5 + ... ))))).
%F A181122 The slowly converging series representation Sum_{n >= 0} (-1)^n/(5*n + 1) for the constant can be accelerated to give the following faster converging series:
%F A181122 1/2 + (5/2)*Sum_{n >= 0} (-1)^n/((5*n + 1)(5*n + 6)) and
%F A181122 17/24 + (25/2)*Sum_{n >= 0} (-1)^n/((5*n + 1)(5*n + 6)*(5*n + 11)).
%F A181122 These two series are the cases r = 1 and r = 2 of the general result: for r >= 0, the constant equals
%F A181122 C(r) + ((5/2)^r)*r!*Sum_{n >= 0} (-1)^n/((5*n + 1)*(5*n + 6)*...*(5*n + 5*r + 1)), where C(r) is the rational number (1/2)*Sum_{k = 0..r-1} (5/2)^k*k!/(1*6*11*...*(5*k + 1)). The general result can be proved by the WZ method as described in Wilf. (End)
%F A181122 From _Peter Bala_, Mar 03 2024: (Start)
%F A181122 Equals hypergeom([1/5, 1], [6/5], -1).
%F A181122 Gauss's continued fraction: 1/(1 + 1^2/(6 + 5^2/(11 + 6^2/(16 + 10^2/(21 + 11^2/(26 + 15^2/(31 + 16^2/(36 + 20^2/(41 + 21^2/(46 + ... )))))))))). (End)
%e A181122 0.88831357265178863804075522702037934627811083077545817120597...
%p A181122 (int(1/(1+x^5),x=0..1));
%p A181122 evalf(LerchPhi(-1,1,1/5)/5) ; # _R. J. Mathar_, Oct 16 2011
%t A181122 (Sqrt[8 + 8/Sqrt[5]]*Pi + 2*Sqrt[5]*ArcCoth[3/Sqrt[5]] + Log[16])/20 // RealDigits[#, 10, 105]& // First (* _Jean-François Alcover_, Feb 13 2013 *)
%o A181122 (PARI)
%o A181122 default(realprecision, 106);
%o A181122 eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+1)))), "3..-2")) \\ _Gheorghe Coserea_, Oct 06 2015
%Y A181122 Cf. A002162, A003881, A024396, A113476, A181048, A181049, A181122, A193534, A262246.
%K A181122 cons,nonn
%O A181122 0,1
%A A181122 _Jonathan D. B. Hodgson_, Oct 05 2010