Let $a$ real number and let $$ a_n=\prod_{k=1}^n\sin(ak).$$ How can I evaluate $$\lim_{n\to+\infty} a_n$$ If $a$ is rational the product is definitely constant but I don’t know how to solve the limit if $a$ is irrational.
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1It may help that the limit of $a_n$ exists only if $\sin(a n)\to 1$. – Giulio R Apr 15 '21 at 14:39
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2@Giulio Actually, $a_n$ approaches the limit of $0$ (except when $a$ is a multiple of $\pi$) – Ben Grossmann Apr 15 '21 at 14:40
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1Ah right :) so the limit is always zero (there are infinitely many factors $<1/2$) – Giulio R Apr 15 '21 at 14:42
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1@Tyrion What do you mean by "the product is definitely constant"? – Ben Grossmann Apr 15 '21 at 14:43
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1@BenGrossmann The limit is zero even for $a$ a multiple of $\pi$. – Vishu Apr 15 '21 at 14:45
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@Tavish You're right; I was thinking of cosine – Ben Grossmann Apr 15 '21 at 14:53
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There are infinitely many factors which are smaller than 1/2 (in absolute value) in the product. This implies that the product is zero.
About the first claim, you may see:
Sine function dense in $[-1,1]$
In general the sequence $\exp(i a n)$ is uniform on the unit circle for all $a$ irrational to $\pi$, yielding the claim.
When $a$ is rational to $\pi$ we do not need this density argument, however the limit is anyway zero.
EDIT: as pointed out in the comment below by Barry Cipra, we need to observe that every factor is smaller than 1, and infinitely many of them are actually smaller than 1/2, and thus the infinite product converges to 0.
Giulio R
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1To be hyperprecise, it's the fact that infinitely many terms are less than $1/2$ (in absolute value) and the fact that all terms are less than or equal to $1$ that implies the limit is $0$. E.g., if the product were taken over $(3/2)\sin(ak)$ instead of just $\sin(ak)$, we'd have infinitely many terms less than $3/4$ in absolute value, but that would not suffice to conclude the product tends to $0$. – Barry Cipra Apr 15 '21 at 16:19