A216992 Decimal expansion of Sum_{n = 1, ..., infinity } 1/n^(2^n).

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

Offset: 1

Author

Nicolas M. Perrault, Sep 21 2012

Keywords

• cons
• easy
• nonn

Comments

The sum converges very quickly and therefore just a few summands are quite enough to get the value accurate to hundreds of decimal places. For example, 1/10^(2^10) = 10^(-1024), meaning that the impact of n = 10 on the sum can't be seen among the first thousand decimal digits. - Alonso del Arte, Sep 21 2012

Examples

			1.0626524160231065162343119079497327861...
		

Crossrefs

Cf. A097547.

Programs

Maple

evalf(sum(1/n^(2^n), n=1..infinity), 140);  # Alois P. Heinz, Sep 29 2023

Mathematica

RealDigits[Sum[1/n^(2^n), {n, 10}], 10, 105][[1]] (* T. D. Noe, Sep 21 2012 *)

PARI

suminf(n=1,1/n^2^n) \\ Charles R Greathouse IV, Apr 21 2016