OFFSET
0,1
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.
LINKS
Daniel Hoyt, Computing the limiting difference between the sum and integral of x^(1/x), 2022, pp. 1-4.
Wikipedia, Stieltjes Constants.
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)
EXAMPLE
0.98854960114226875064475410833997126442199868380...
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 *)
PROG
(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])
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Daniel Hoyt, Jun 16 2023
STATUS
approved