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Decimal expansion of Sum_{k>=1} 1/((2k-1)^2(2k+1)^2) = (Pi^2-8)/16.
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%I #14 Aug 20 2024 02:06:07

%S 1,1,6,8,5,0,2,7,5,0,6,8,0,8,4,9,1,3,6,7,7,1,5,5,6,8,7,4,9,2,2,5,9,4,

%T 4,5,9,5,7,1,0,6,2,1,2,9,5,2,5,4,9,4,1,4,1,5,0,8,3,4,3,3,6,0,1,3,7,5,

%U 2,8,0,1,4,0,1,2,0,0,3,2,7,6,8,7,6,1,0,8,3,7,7,3,2,4,0,9,5,1,4,4,8,9,0,0,1

%N Decimal expansion of Sum_{k>=1} 1/((2k-1)^2(2k+1)^2) = (Pi^2-8)/16.

%D Erwin Kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley and Sons, Inc., NJ, 2006, page 506.

%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>.

%F Equals (A111003-1)/2. - _Hugo Pfoertner_, Aug 20 2024

%e 0.116850275068084913677155687492259445957106212952549414150834336...

%t RealDigits[Sum[1/((2k - 1)^2(2k + 1)^2), {k, Infinity}], 10, 111][[1]]

%o (PARI) (Pi^2-8)/16 \\ _Charles R Greathouse IV_, Sep 30 2022

%Y Cf. A000796, A002388, A069075, A111003.

%K cons,nonn

%O 0,3

%A _Robert G. Wilson v_, Sep 27 2006