If k = 1+1/x, show that for large x, Zeta(k) = x?

Question

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Given the infinite series Σ (1/n^k) where n = 1 to infinite, if k = 1 + 1/x, for large x show that the sum asymptotically equals x. In the case of x = infinity, then we have the classic series Σ (1/n) which does not converge.

Answer 1

For any k>0, 1/(n+1)^k ≤ 1/ceiling (n)^k ≤ 1/n^k. Therefore:

[1, ∞]∫1/(n+1)^k dn ≤ [1, ∞]∫1/ceiling (n)^k dn ≤ [1, ∞]∫1/n^k dn.

However for any integer a, 1/ceiling (n)^k = 1/(a+1)^k almost everywhere over the region n∈[a, a+1]. So, it follows that [a, a+1]∫1/ceiling (n)^k dn = 1/(a+1)^k, and therefore that [1, ∞]∫1/ceiling (n)^k dn = [n=1, ∞]∑1/(n+1)^k = [n=2, ∞]∑1/n^k = ζ(k) - 1.Therefore:

[1, ∞]∫1/(n+1)^k dn ≤ ζ(k) - 1 ≤ [1, ∞]∫1/n^k dn.

Substituting k=1+1/x:

[1, ∞]∫1/(n+1)^(1+1/x) dn ≤ ζ(1+1/x) - 1 ≤ [1, ∞]∫1/n^(1+1/x) dn

Resolving the integrals:

x(n+1)^(-1/x) |[1, ∞] ≤ ζ(1+1/x) - 1 ≤ xn^(-1/x) |[1, ∞]
2^(-1/x) * x ≤ ζ(1+1/x) - 1 ≤ x

Further algebra yields:

2^(-1/x) * x + 1 ≤ ζ(1+1/x) ≤ x + 1
2^(-1/x) + 1/x ≤ ζ(1+1/x)/x ≤ 1 + 1/x

Now, [x→∞]lim 2^(-1/x) + 1/x = 1 and [x→∞]lim 1 + 1/x = 1, so by the squeeze theorem, [x→∞]lim ζ(1+1/x)/x = 1, so ζ(1+1/x) ~ x.

Actually, a stronger statement is possible. Recall that:

2^(-1/x) * x ≤ ζ(1+1/x) - 1 ≤ x

So simple algebra yields:

2^(-1/x)*x - x + 1 ≤ ζ(1+1/x) - x ≤ 1

Obviously, [x→∞]lim 1 = 1, and

[x→∞]lim 2^(-1/x)*x - x + 1
1 + [x→∞]lim x*(2^(-1/x)-1)
1 + [x→∞]lim (2^(-1/x)-1)/(1/x)
1 + [x→∞]lim (e^(-ln 2/x)-1)/(1/x)
1 + [x→∞]lim (e^(-ln 2/x) * ln 2/x²)/(-1/x²) (using L'hopital's rule)
1 + [x→∞]lim - ln 2 e^(-ln 2/x)
1-ln 2

So 1-ln 2 ≤ [x→∞]lim inf ζ(1+1/x) - x ≤ [x→∞]lim sup ζ(1+1/x) - x ≤ 1 -- that is, the limiting difference between ζ(1+1/x) and x is somewhere between 1-ln 2 and 1. I suspect that it may be related to the Euler-Mascheroni constant, but I don't know how to pin it down any further.