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A350862 Decimal expansion of Sum_{k>=1} (k^(1/k^(1 + 1/1111)) - 1).

%I A350862 #28 Feb 06 2025 14:25:20
%S A350862 1,2,3,4,3,2,1,9,8,2,8,1,3,8,9,0,9,3,3,3,6,8,6,4,2,4,4,0,0,4,8,8,7,7,
%T A350862 4,8,6,8,2,6,9,1,2,5,8,7,7,1,5,4,8,3,8,1,2,6,2,3,5,0,2,6,6,6,4,0,7,4,
%U A350862 2,2,6,9,9,0,2,7,0,3,0,1,1,3,8,2,7,7,9
%N A350862 Decimal expansion of Sum_{k>=1} (k^(1/k^(1 + 1/1111)) - 1).
%C A350862 If x is a whole number greater than 1, the Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(x))) - 1) = x + C, where C is a constant less than 1.
%C A350862 The above relation was tested for all 1 < x < 10^7.
%C A350862 If x = 1, the sum is A329117.
%C A350862 This sum demonstrates this relationship: setting sqrt(x) = 1111 generates the sum 1111^2 + C or 1234321 + C. Another example would be Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(1729))) - 1) = 1729.84841430674....
%C A350862 Evaluating the sum at larger x values converges slower and slower. Monotonically changing extrapolation methods such as Richardson's Extrapolation must be used to compute these values.
%C A350862 Since the output (x + C) will be the square of the input (sqrt(x)) plus a constant less than 1, this implies that Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(x))) - 1) diverges as x tends to infinity, or simplified to Sum_{k>=1} (k^(1/k) - 1).
%H A350862 Daniel Hoyt, <a href="/A351885/a351885.pdf">Computing the limiting difference between the sum and integral of x^(1/x)</a>, 2022, pp. 3-4.
%e A350862 1234321.98281389093336864244004887748682691258771548...
%t A350862 digits = 120; d = 1; j = 1; s = 0; While[Abs[d] > 10^(-digits - 5), d = (-1)^j/j!*Derivative[j][Zeta][(1 + 1/1111)*j]; s += d; j++]; RealDigits[s, 10, 120][[1]] (* _Vaclav Kotesovec_, Jun 18 2023 *)
%o A350862 (PARI) sumpos(k=1, k^(1/(k^(1 + 1/1111))) - 1)
%Y A350862 Cf. A329117, A308511.
%K A350862 nonn,cons
%O A350862 7,2
%A A350862 _Daniel Hoyt_, Jan 19 2022