Are any closed form solutions / upper bounds known for this product?

Question

I'm trying to get a closed form solution for the following product:

$$ \prod_{i=1}^t\left(1-\frac{a}{i^\beta}\right) $$

where $\beta\in[1,2]$ and $a>0$. In my application, an upper bound that converges to zero as $t\to\infty$ would also work.

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Edit:

Extending the method linked by @DietrichBurde in the comments, we get

\begin{align} \prod_{i=1}^t\left(1-\frac{a}{i^\beta}\right) &= \prod_{i=1}^t\frac{i^\beta-a}{i^\beta} = \frac{\prod_{i=1}^t i^\beta-a}{\prod_{i=1}^ti^\beta} = \frac{\prod_{i=1}^t i^\beta-a}{\left(\prod_{i=1}^ti\right)^\beta} = \frac{\prod_{i=1}^t i^\beta-a}{\Gamma\left(t+1\right)^\beta} \end{align} All that remains is to evaluate the numerator. For $\beta=1$, we have $$ \prod_{i=1}^t (i-a) = \frac{\Gamma(t-a+1)}{\Gamma(a+1)} $$ and so $$ \prod_{i=1}^t\left(1-\frac{a}{i}\right) = \frac{\Gamma(t-a+1)}{\Gamma(a+1)\Gamma(t+1)} $$ as in the linked answer.

For $\beta=2$, we can factorize to get \begin{align} \prod_{i=1}^t (i^2-a) &= \prod_{i=1}^t (i-\sqrt{a})(i+\sqrt{a}) \\&= \left(\prod_{i=1}^t (i-\sqrt{a})\right)\left(\prod_{i=1}^t (i+\sqrt{a})\right) \\&= \frac{\Gamma(t-\sqrt{a}+1)\Gamma(t+\sqrt{a}+1)}{\Gamma(-\sqrt{a}+1)\Gamma(\sqrt{a}+1)} \\&= \frac{\Gamma(t-\sqrt{a}+1)\Gamma(t+\sqrt{a}+1)}{\Gamma(-\sqrt{a}+1)\Gamma(\sqrt{a})(a+1)} \\&= \frac{\sin(\pi \sqrt a)}{\pi(a+1)}\Gamma(t-\sqrt{a}+1)\Gamma(t+\sqrt{a}+1) \end{align} where the last line follows from Eurler's reflection formula. Therefore, $$ \prod_{i=1}^t\left(1-\frac{a}{i^2}\right) = \frac{\sin(\pi \sqrt a)\Gamma(t-\sqrt{a}+1)\Gamma(t+\sqrt{a}+1)}{\pi(a+1)\Gamma(t+1)^2}. $$ But now I'm not sure what to do with fractional $\beta$. In particular, how can we rewrite $\prod_{i=1}^t (i^\beta -a)$ in a more useful form?

Answer 1

For $\beta > 1$ and $a \in (0,1)$, an easy to derive upper bound is $$\begin{align}\prod_{k=1}^p\left(1 - \frac{a}{k^\beta}\right) &\le \exp\left(-\sum\limits_{k=1}^p\frac{a}{k^\beta}\right) = \exp\left(-a\zeta(\beta) + a\sum\limits_{k=p+1}^\infty \frac{1}{k^\beta}\right)\\ &\le \exp\left(-a\zeta(\beta)+a\int_p^\infty \frac{dx}{x^\beta}\right) = \exp\left(-a\zeta(\beta)+ \frac{a}{\beta-1}p^{1-\beta}\right) \end{align} $$ If this upper bound is not tight enough, one can use the fact all coefficients in following expansion is negative $$\log(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots$$ Repeat essentially the same argument, we find for any positive integer $m$,

$$\prod_{k=1}^p \left(1 - \frac{a}{k^\beta}\right) \le \exp\left( -\sum\limits_{k=1}^p \sum\limits_{n=1}^m \frac{1}{n}\frac{a^n}{k^{n\beta}}\right) \le \exp\left[-\sum\limits_{n=1}^m \frac{a^n}{n} \left(\zeta(n\beta) - \frac{1}{n\beta-1}p^{1-n\beta}\right)\right] $$

Update

For $\beta \in (0,1]$ and $a \in (0,1)$, we have $$\begin{align}\prod_{k=1}^p\left(1 - \frac{a}{k^\beta}\right) &\le \exp\left(-\sum\limits_{k=1}^p\frac{a}{k^\beta}\right) = \exp\left[-a\left( 1 + \sum\limits_{k=2}^p\frac{1}{k^\beta}\right)\right]\\ &\le \exp\left[ -a\left(1 + \int_1^p \frac{dx}{x^\beta}\right)\right]\\ &= \begin{cases} \exp\left[ -a\left(1 + \frac{p^{1-\beta} - 1}{1-\beta}\right)\right], & \beta < 1\\ \exp(-a(1 + \log p)) = (pe)^{-a}, &\beta = 1 \end{cases} \end{align} $$ In both cases, the bound on RHS decrease to $0$ as $p$ increases. By squeezing, we find the product converges to $0$ as $p \to \infty$.