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Decimal expansion of 1/2 + 1/3 + 1/5 + ... + 1/(2^n + 1) + ...
6

%I #29 Jan 30 2022 15:58:11

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

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

%U 7,5,6,7,9,7,6,6,0,0,7,0,8,5,0,6,1,9

%N Decimal expansion of 1/2 + 1/3 + 1/5 + ... + 1/(2^n + 1) + ...

%H Jinyuan Wang, <a href="/A323482/b323482.txt">Table of n, a(n) for n = 1..1000</a>

%F From _Amiram Eldar_, Jun 30 2020: (Start)

%F Equals 1/2 + Sum_{k>=1} (-1)^(k+1)/(2^k-1)

%F Equals Sum_{k>=1} (mu(k) - (-1)^k)/(2^k-1), where mu is the Möbius function (A008683).

%F Equals (1 + A179951)/2. (End)

%F Equals (1/2) + Sum_{n >= 1} (-1)^(n+1)*((4^n + 1)/(4^n - 1))*(1/2)^(n^2). The first 18 terms of the series gives the constant correct to more than 100 decimal places. - _Peter Bala_, Jan 28 2022

%e 1.2644997803484442091913...

%p evalf((1/2) + add( (-1)^(n+1)*((4^n + 1)/(4^n - 1))*(1/2)^(n^2), n = 1..18), 100); # _Peter Bala_, Jan 28 2022

%t s = Sum[1/(2^k + 1), {k, 0, Infinity}]

%t r = N[Re[s], 200]

%t RealDigits[r][[1]]

%o (PARI) suminf(k=0, 1/(2^k+1)) \\ _Michel Marcus_, Jan 15 2019

%Y Cf. A000051, A065442, A008683, A179951.

%K nonn,cons

%O 1,2

%A _Clark Kimberling_, Jan 15 2019