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Explicit tail bound for the log-Gamma product series

Statement

For $Re z > 0$ and integer $N > 2 ∣ z ∣$ , the tail in the series for $ℓ_{+} (z)$ after its first $N$ summands satisfies

$∣ \sum_{n = N + 1}^{\infty} (\frac{z}{n} - Log (1 + \frac{z}{n})) ∣ \leq \frac{∣ z ∣^{2}}{2 N (1 - ∣ z ∣ / (N + 1))} .$

Facts & Assumptions

Given: $Re z > 0$ and an integer $N > 2 ∣ z ∣$ .

[F1]

Log-Gamma branch on the right half-plane: the summands are $z / n - Log (1 + z / n)$ .

[F2]

The principal logarithm is the normalised holomorphic branch on the slit plane: ${Log}^{'} (w) = 1 / w$ on the slit plane.

[F3]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : the derivative of a continuously differentiable complex-valued path integrates to its endpoint difference, by applying the real theorem to each component.

[F4]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuous function on $[0, 1]$ is Riemann integrable.

Proof

1.1givenF1F2F3F4algebra

Put $w = z / n$ , so $Re w > 0$ . The integrand in the path formula is continuous and integrable by [F4]. By [F2], [F3], and the chain rule, $w - Log (1 + w) = \int_{0}^{1} \frac{t w^{2}}{1 + t w} d t .$ Since $∣ 1 + t w ∣^{2} = 1 + 2 t Re w + t^{2} ∣ w ∣^{2} \geq 1$ , this summand has modulus at most $∣ z ∣^{2} / (2 n^{2})$ .

2.1givenstep 1.1algebra∎

The decreasing function $t \mapsto t^{- 2}$ satisfies $\sum_{n = N + 1}^{\infty} n^{- 2} \leq \int_{N}^{\infty} t^{- 2} d t = 1 / N$ . Summing step 1.1 gives $∣ z ∣^{2} / (2 N)$ . Since $0 < 1 - ∣ z ∣ / (N + 1) < 1$ under $N > 2 ∣ z ∣$ , this is no larger than the stated bound.

Depends on

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Sources

DLMF, §5.8(ii), equation 5.8.2 (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§2 and 3 (standard reference, not scraped)