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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 $0$?

lzrus
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    We have $$ \lim_{n\rightarrow \infty} \left( 1- \frac{\alpha}{n} \right)^n = e^{-\alpha} \neq 0.$$ – Severin Schraven Jan 31 '23 at 04:49
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    Positive precludes zero, so you mean nonnegative not positive. Why would $\alpha$ positive imply this though? For $\alpha=1.5$ for example the product on the LHS is always negative. Anyway, ignoring all factors with $n<\alpha$, we can take the logarithm of the truncated product and use $\log(1+x)=x+{\cal O}(x^2)$ for $|x|<1$ to see the product must diverge to $0$, so at least your conclusion is true. – anon Jan 31 '23 at 04:51
  • To elaborate a bit on runway44's comment. First show that $L=\lim_{n\rightarrow \infty} \prod_{j=1}^n \left( 1-\frac{\alpha}{n}\right)$ exists (use monotonicity after a certain index and boundedness). If $L\neq 0$, then pick $s\in {\pm 1}$ such that $sL>0$ and observe that $$\ln(sL) = \lim_{n\rightarrow \infty} \ln \left( s \prod_{j=1}^n \left(1-\frac{\alpha}{n}\right) \right) = \ln\left( s\prod_{j=1}^N \left(1-\frac{\alpha}{n}\right) \right) + \sum_{j=N+1}^n \ln\left(1-\frac{\alpha}{j}\right).$$ Now use runway's comment to show that the series diverges which gives the desired contradiction. – Severin Schraven Jan 31 '23 at 05:18
  • Of course the factors in the formula in my previous comment should be $\left( 1-\frac{\alpha}{j}\right)$, not $\left( 1-\frac{\alpha}{n}\right)$. – Severin Schraven Jan 31 '23 at 06:13
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    $$ \frac{1}{{n!}}\prod\limits_{i = 1}^n {(i - \alpha )} = \frac{1}{{\Gamma (1 - \alpha )}}\frac{{\Gamma (n + 1 - \alpha )}}{{\Gamma (n + 1)}} \sim \frac{1}{{\Gamma (1 - \alpha )}}\frac{1}{{(n + 1)^\alpha }} $$ – Gary Jan 31 '23 at 07:41
  • Handle the case $\alpha\in\Bbb N$ separately. – J.G. Jan 31 '23 at 08:22

2 Answers2

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

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

Anne Bauval
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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 $\alpha >0$.

grand_chat
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