Abstract
Let H_{n} = sum_{r=1}^{n} 1/r and H_{n}(x) = sum_{r=1}^{n} 1/(r+x). Let psi(x) denote the digamma function. It is shown that H_{n}(x) + psi(x+1) is approximated by frac{1}{2}log f(n+x), where f(x) = x^{2} + x + frac{1}{3}, with error term of order (n+x)^{-5}. The cases x = 0 and n = 0 equate to estimates for H_{n} - gamma and psi(x+1) itself. The result is applied to determine exact bounds for a remainder term occurring in the Dirichlet divisor problem.
Highlights
Introduction and summary of resultsWrite Hn for the harmonic sum n r=1 1 rThe following well-known estimation can be established by an Euler–Maclaurin summation, or by the logarithmic and binomial series: Hn – γ = log n + 2n – 12n2 + rn, (1)where γ is Euler’s constant and < rn ≤ Since log(n ) log n
One natural extension is the replacement of Hn – γ by ψ(x + 1), where ψ(x) is the digamma function Γ (x)/Γ (x), since Hn – γ = ψ(n + 1)
Theorem 1 has a rather surprising application to an expression that arises in the Dirichlet divisor problem
Summary
Numerous more recent articles have developed this process further. Chen and Mortici [3] show that. Further variations and extensions are given, for example, in [7] and [4], and in other references listed in these papers. One natural extension is the replacement of Hn – γ by ψ(x + 1), where ψ(x) is the digamma function Γ (x)/Γ (x), since Hn – γ = ψ(n + 1). To identify the limit of Hn(x) – log n, recall Euler’s limit formula for the Gamma function: this can be written as Γ (x + 1) = limn→∞ Gn(x + 1), where nx(n + 1)! (x + n) , from which it follows that limn→∞[Hn(x) – log n] = –ψ(x + 1).
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