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Finite harmonic enclosure for Euler's constant

Statement

For each integer $M \geq 1$ , put $H_{M} = \sum_{k = 1}^{M} 1 / k$ . Euler's constant $γ$ satisfies $0 \leq H_{M} - \log M - γ \leq \frac{1}{M} .$ Thus $[H_{M} - \log M - 1 / M, H_{M} - \log M]$ is a finite enclosure for $γ$ .

Facts & Assumptions

Given: An integer $M \geq 1$ .

[F1]

The Euler–Mascheroni constant and the harmonic asymptotic: $γ_{n} = H_{n} - \log n$ decreases to $γ$ .

[F2]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: $\log (n + 1) - \log n = \int_{n}^{n + 1} d t / t$ for $n \geq 1$ .

Proof

1.1F1F2algebra

For each $n \geq M$ , [F1] and [F2] give $γ_{n} - γ_{n + 1} = \int_{n}^{n + 1} \frac{d t}{t} - \frac{1}{n + 1} = \int_{n}^{n + 1} (\frac{1}{t} - \frac{1}{n + 1}) d t .$ The integrand is nonnegative and at most $1 / n - 1 / (n + 1)$ on this interval. Therefore $0 \leq γ_{n} - γ_{n + 1} \leq \frac{1}{n} - \frac{1}{n + 1} .$

2.1F1step 1.1algebra∎

Summing step 1.1 for $n = M, \dots, K$ telescopes to $0 \leq γ_{M} - γ_{K + 1} \leq \frac{1}{M} - \frac{1}{K + 1} .$ Let $K \to \infty$ and use $γ_{K + 1} \to γ$ from [F1]. Since $γ_{M} = H_{M} - \log M$ , the claimed two-sided bound follows.

Depends on

Used by

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Sources

DLMF, §5.2, Gamma and psi functions (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§2 and 3 (standard reference, not scraped)