Limit of $\frac{1}{n!}\prod_{i=1}^{n}(i-\alpha)$

Question

Suppose that $\alpha > 0$. Show that $$\lim_{n\to\infty}\left(1-\frac{\alpha}{1}\right)\left(1-\frac{\alpha}{2}\right)\ldots\left(1-\frac{\alpha}{n}\right)=0$$ Here’s what I’ve tried so far, $$ \begin{align} \lim_{n\to\infty}\left(1-\frac{\alpha}{1}\right)\left(1-\frac{\alpha}{2}\right)\ldots\left(1-\frac{\alpha}{n}\right) &\leq \lim_{n\to \infty}\left(1-\frac{\alpha}{n}\right)^n\\ &=0 \end{align} $$ Since $\alpha$ is positive, then the LHS limit is positive as well. As such, can we simply conclude that LHS limit is $0$?

Answer 1

$\ln\prod_{n>\alpha}\left(1-\frac\alpha n\right)=\sum_{n>\alpha}\ln\left(1-\frac\alpha n\right)=-\infty,$ because $\ln(1+x)\sim_{x\to0}x$ and the harmonic series diverges.

Therefore, $\prod_{n>\alpha}\left(1-\frac\alpha n\right)=0.$

Answer 2

Hint: Use the inequality $1+x\le e^x$ to see $$ \prod_{i=1}^n\left( 1 -\frac\alpha i \right) \le \prod_{i=1}^n\exp\left( - \frac\alpha i\right)=\exp\left(-\alpha \left[1 + \frac12+\frac13+\cdots+\frac 1n\right]\right). $$ Next, use the inequality $$1 + \frac12+\frac13+\cdots+\frac 1n > \log n.$$ Notice where you use the fact that $\alpha >0$.