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For what values of $a$ does $P=\prod\limits_{k=1}^n a|\sin{k}|\to\infty$ as $n\to\infty$ ?

Experimenting on desmos, it seemed that if $a>2$ then $P\to\infty$, but some strange cases like $\prod\limits_{k=1}^{120000} 2.0001|\sin{k}|\approx 4\times10^{-17}$ made me doubt it.

Either there exists a critical value for $a$ such that $P\to\infty$, or $P\not\to\infty$ for all $a$. Either way, I think it's astounding.

Dan
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  • So what if one considers $a$ s.t. one gets a product that diverges? A product diverges when ...? – mavavilj Sep 25 '22 at 09:41
  • @mavavilj Does such an $a$ exist? – Dan Sep 25 '22 at 09:44
  • If $u_n = \prod_{k=1}^n |\sin k|$, one could search the $\limsup \frac{\log |\log u_n|}{\log n}$ first. – Gribouillis Sep 25 '22 at 09:50
  • https://math.stackexchange.com/questions/381254/the-infinite-product-of-the-sines-of-all-positive-integers-is-zero, https://en.wikipedia.org/wiki/Infinite_product – mavavilj Sep 25 '22 at 10:16
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    https://math.stackexchange.com/questions/3356735/divergence-of-prod-n-1-infty-a-sinn-for-a1-to-0-or-infty – mavavilj Sep 26 '22 at 08:39

1 Answers1

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This is only an approach of the problem. We have $P_n = a^n u_n$ where $u_n = \prod_{k=1}^n |\sin k|$. Clearly, $u_n$ converges to $0$, hence $\log u_n$ tends to $-\infty$.

Proposition Let $\ell = \limsup \frac{\log |\log u_n|}{\log n}$. If $\ell > 1$ then no such $a$ exists.

Proof: if $\ell > 1$, let $1 <\alpha< \ell$. For an infinite number of $n$ we have $|\log u_n| \ge n^\alpha$, hence $\log u_n\le -n^\alpha$, hence $\log P_n\le n \log a - n^\alpha$ which tends to $-\infty$ for this subsequence of ns.

Now have you tried this $\limsup$ numerically?

Second question Can we expect that for large $n$ \begin{equation} \frac{1}{n}\sum_{k=1}^n \log|\sin k| \approx \frac{1}{\pi}\int_0^\pi\log|\sin\theta| d \theta\quad? \end{equation}

Gribouillis
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  • Why is it clear that $u_n$ tends to $0$? – Bruno Krams Sep 25 '22 at 10:38
  • @BrunoKrams It is clear because $u_n$ is a product of numbers in $(0, 1)$ which already means that it decreases, ie $u_{n}>u_{n+1}>0$. Furthermore, we have $\sin n < \frac{1}{2}$ for an infinite number of $n$ s. Thus the product is at least divided by more than 2 an infinite number of times (other arguments are possible). – Gribouillis Sep 25 '22 at 10:42
  • @Gribouillis I tested the limsup on desmos, but it quits working beyond n=1064, at which the limsup is $0.9476....$. $\frac{1}{n}\sum_{k=1}^n \ln|\sin k|\to-\ln2$ and $\frac{1}{\pi}\int_0^\pi\log|\sin\theta| d \theta=-\ln{2}$. – Dan Sep 25 '22 at 11:03
  • @Dan this looks like a consequence of ergodicity of the translation of 1 radian on a circle of length $2\pi$. Can you prove the limits? This would already indicate that $\log u_n\sim \log 2^{-n}$. Also the limsup above would be $\ell=1$. – Gribouillis Sep 25 '22 at 11:47
  • @Gribouillis I think your "second question" shows that $P\to\infty$ if and only if $a>2$. Fair enough. If $a$ is just slightly bigger than $2$, then $P$ certainly takes a while to get going on its way to $\infty$. For example, if $a=2.01$, then from $n=1$ to $n=6500$ or so, $P$ seems to stay small, dipping above and below $1$ many times. – Dan Sep 25 '22 at 11:55
  • What's this second question about? Integral test? – mavavilj Sep 26 '22 at 08:30
  • @mavavilj No, it is an ergodic result similar to Birkhoff's ergodic theorem, using the ergodic translation $T(x) = x+1$ in $\mathbb{R}/2\pi{\mathbb Z}$. In fact the result here is not exactly implied by this theorem but it remains true despite the fact that the function $\log |\sin \theta|$ is not continuous in $[0, 2\pi]$. Try to provide a complete proof, which is currently missing here! – Gribouillis Sep 26 '22 at 12:26
  • It needs some elaboration I think. Compared to https://math.stackexchange.com/a/3356772/248602 this answer seems quite technical to me. – mavavilj Sep 26 '22 at 18:00