# Approximating $\prod_{i=1}^{n-1} (1-ai)$ for large $n$

I have a function of the form:

$f(n) = \prod_{i=1}^{n-1} (1-ai)$

Here, $a \geq 0$ and $(a*i) < 1$. For $n > 10^5$ or $10^6$, what is the best possible analytic approximation for $f(n)$ that will allow me compute values for the function with reasonable computational resources? What bounds on the error of the approximation can I expect?

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What is the role of $i$ in the product? – András Bátkai Aug 15 '12 at 17:09
@András Bátkai, that was incredibly sloppy, sorry about that. I've fixed f(x). – Roger S. Aug 15 '12 at 17:11
You could group terms in pairs to get terms like (1 - (n+1)a + ba^2). For small values of a this might help you with your error estimates. Gerhard "Ask Me About System Design" Paseman, 2012.08.15 – Gerhard Paseman Aug 15 '12 at 17:15
@Gerhard Paseman, thanks for the good suggestion! I've been trying things a bit like that, but the trouble is that $a$ isn't necessarily small, and I'm looking for some error bounds for the approximation. – Roger S. Aug 15 '12 at 17:40

$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$ If $a n < 1$, you'll want to use the reflection principle $$\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}}$$ so $$f(n) = \dfrac{a^{n-1} \Gamma(1/a)}{\Gamma(1-n+1/a)}$$
Use Stirling's series for the asymptotic approximation of $\ln(\Gamma(1-n+ 1/a))$, or one of its variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation