A363704 Decimal expansion of lim_{x -> infinity} ((Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1)) - x^2).

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

Offset: 0

Author

Daniel Hoyt, Jun 16 2023

Keywords

• nonn
• cons

Comments

That this constant is less than one allows Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1) = floor(x^2), when x is the square root of any natural number greater than 1.

The limit converges slowly.

Examples

			0.98854960114226875064475410833997126442199868380...
		

Links

• Daniel Hoyt, Computing the limiting difference between the sum and integral of x^(1/x), 2022, pp. 1-4.
• Wikipedia, Stieltjes Constants.

Crossrefs

Cf. A098572, A350862, A329117, A351885, A001620, A363716.

Programs

Mathematica

digits = 120; d = 1; j = 2; s = StieltjesGamma[1]; While[Abs[d] > 10^(-digits - 5), d = (-1)^j / j! * Derivative[j][Zeta][j]; s += d; j++]; RealDigits[s, 10, 120][[1]] (* Vaclav Kotesovec, Jun 17 2023 *)

Python

# Gives 14 correct digits
from mpmath import stieltjes,fac
def limgen(n):
    terms = []
    for y in range(3, n):
        for x in range(y, n):
            terms.append((((-1)**y)*stieltjes(x)*(x-(y-1))**(y-2))/(fac(x-(y-2))*fac(y-2)))
    n,o_sum = 2,0
    while True:
        n_term = 1/((n-1)**(n+1))
        n_sum = o_sum + n_term
        if o_sum == n_sum:
            break
        o_sum = n_sum
        n += 1
    return sum(terms) + 0.5 - stieltjes(0) + n_sum
print(str(limgen(60))[:-1])

Formula

Equals 1/2 - A001620 + Sum_{k>=2} (1/(k-1)^(k+1)) + Sum_{k>=3} Sum_{n>=k} (((-1)^k)*Stieltjes(n)*(n-k+1)^(k-2))/((n-k+2)!*(k-2)!).

From Vaclav Kotesovec, Jun 17 2023: (Start)

Equals lim_{n->oo} (Sum_{m=1..n} m^(1/m)) - n - log(n)^2/2.

Equals sg1 + Sum_{k>=2} (-1)^k / k! * k-th derivative of zeta(k), where sg1 is the first Stieltjes constant (see A082633). (End)