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.

  • $\begingroup$ "Solve for" isn't the right expression here. I'd say "evalutate". $\endgroup$ – Michael Hardy Dec 20 '11 at 18:08
  • 1
    $\begingroup$ 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

$$ \prod_{k=1}^{n}\left(1 - \frac{1}{ak}\right)=\frac{\Gamma(n+1-\frac1a)}{\Gamma(1-\frac1a)\Gamma(n+1)}=\frac1{n\mathrm B(n,1-\frac1a)} $$


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

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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.