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A simple method for evaluating a product is term cancellation. For example, the product

$$\begin{align*} \prod_{k=2}^{n}\left(1-\frac{1}{k}\right)&=\prod_{k=2}^{n}\left(\frac{k-1}{k}\right)\\ &=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdots\frac{n-1}{n} = \frac{1}{n} \end{align*}$$

is done via a telescoping argument. However, if we take a product that is just a bit more complicated, say

$$\prod_{k=1}^{n}\left(1 - \frac{1}{ak}\right)$$

for some $0,1 \neq a \in \mathbb{R}$, the argument immediately breaks down. I am interested in whether there exists a closed form solution for the general product given above.

This question is in part inspired by my attempt to prove the asymptotic bound here.

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  • $\begingroup$ "Solve for" isn't the right expression here. I'd say "evalutate". $\endgroup$ – Michael Hardy Dec 20 '11 at 18:08
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    $\begingroup$ Using gamma function will be helpful in both obtaining a closed form and finding an asymptotic formula. $\endgroup$ – Sangchul Lee Dec 20 '11 at 18:09
  • $\begingroup$ Let $b=a^{-1}$, you can see this is a polynomial in $b$: $$\frac {(-1)^n}{n!}\prod_{k=1}^n (b-k)$$ Aside from the sign, this is just the continuation of $b-1\choose n$ to all $b\neq 0$. $\endgroup$ – Thomas Andrews Dec 20 '11 at 18:10
  • $\begingroup$ @MichaelHardy Maybe expand is a better choice? I am not a native English speaker though. $\endgroup$ – AD. Dec 20 '11 at 19:21
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$$ \prod_{k=1}^{n}\left(1 - \frac{1}{ak}\right)=\frac{\Gamma(n+1-\frac1a)}{\Gamma(1-\frac1a)\Gamma(n+1)}=\frac1{n\mathrm B(n,1-\frac1a)} $$

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$$\prod_{k=1}^n\left(1 - \frac1{ak}\right)=\frac{\prod\limits_{k=1}^n \left(k-\frac1{a}\right)}{n!}=\frac{\prod\limits_{k=0}^{n-1} \left(-\frac1{a}+k+1\right)}{n!}=\frac{\left(1-\frac1{a}\right)_n}{n!}$$

where $(a)_n=\prod\limits_{j=0}^{n-1}(a+j)=\frac{\Gamma(a+n)}{\Gamma(a)}$ is the Pochhammer symbol. (Essentially, the second expression in @Did's answer.)

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