Asymptotic series for the reciprocal of the gamma function

Stirling’s asymptotic series for the gamma function is

$\Gamma(z) \sim (2\pi)^{1/2} z^{z - 1/2} e^{-z} \sum_{n=0}^\infty (-1)^n \frac{\gamma_n}{z^n}$

Now suppose you’d like to find an asymptotic series for the function 1/Γ(z).

Since the series for Γ has the form f(z) times an infinite sum, it would make sense to look for a series for 1/Γ of the form 1/f(z) times an infinite sum. The hard part would be finding the new infinite sum. In general the series for function and the series for its reciprocal look nothing alike.

Here’s where we have a pleasant surprise: the coefficients in the series for 1/Γ are exactly the same as the coefficients in the series for Γ, except the signs don’t alternate.

$\frac{1}{\Gamma(z)} \sim (2\pi)^{-1/2} z^{-z +1/2} e^{z} \sum_{n=0}^\infty \frac{\gamma_n}{z^n}$

Illustration

The following is not a proof, but it shows that the result is at least plausible.

Define Γ* to be Γ divided by the term in front of the infinite series:

$\Gamma^*(z) = (2\pi)^{-1/2} z^{-z +1/2} e^{z} \Gamma(z)$

Then the discussion above claims that Γ* and 1/Γ* have the same asymptotic series, except with alternating signs on the coefficients. So if we multiply the first few terms of the series for Γ* and 1/Γ* we expect to get something approximately equal to 1.

Now

$\Gamma^*(z) = 1 + \frac{1}{12z} + \frac{1}{288z^2} - \frac{139}{51840z^3} - \cdots$

and we claim

$\frac{1}{\Gamma^*(z)} = 1 - \frac{1}{12z} + \frac{1}{288z^2} + \frac{139}{51840z^3} - \cdots$

So if we multiply the terms up to third order we expect to get 1 and some terms involving powers of z in the denominator with exponent greater than 3. In fact the product equals

$1 + \frac{571}{1244160 z^4} -\frac{19321}{2687385600 z^6}$

which aligns with our expectations.

Series for the reciprocal of the gamma function

Illustration

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