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The limits I'm trying to solve are:

$$\lim_{n\to \infty}\sin(1)\times\sin(2)\times\sin(3)\times\ldots\times\sin(n)$$ $$\lim_{n\to \infty}n\times\sin(1)\times\sin(2)\times\sin(3)\times\ldots\times\sin(n)$$

For the former limit, my (probably incorrect) solution is that $\sin(1)\times\sin(2)\times\sin(3)\ldots$ are constants, so the limit can be written as

$$\sin(1)\times\sin(2)\times\sin(3)\times\ldots\times\sin(n-1)\cdot \lim_{n\to \infty}\sin(n)$$

and $\lim_{n\to \infty}\sin(n)$ simply does not exist, because $\sin(n)$ does not settle on a single value when ${n\to \infty}$.

Javad Kouhi
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    By the way, your solution doesn't make any sense, because $\sin(1)\sin(2) \cdots \sin(n-1)$ is still a function of $n$. Here's a question: have you tried computing a few small values of this sequence to look for trends? – Erick Wong Aug 17 '15 at 22:06
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    For 1, note each interval $[n\pi-{\pi\over4}, n\pi+{\pi\over4}]$, contains an integer. – David Mitra Aug 17 '15 at 22:16
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    For 2, you could use the first answer here (and may as well use it for 1.). – David Mitra Aug 17 '15 at 22:22
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    Hmm. From my first comment, at least $1/4$ of the terms are bounded by $\sqrt2/2$ in absolute value. You can use this to show the limit is $0$ (but the argument referenced in my second comment is nicer). – David Mitra Aug 17 '15 at 22:37

4 Answers4

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For the first limit, it should be $0$ for the following reason: $\;\{\sin n\mid n\in \mathbf N\}$ is dense in $[-1,1]$. So for any $\varepsilon >0$ there exists $N$ such that $\lvert\, \sin N\,\rvert<\varepsilon$, which implies that for any $n\ge N$, $$\lvert\,\sin 1\cdot \sin 2\cdots \sin N\cdots\sin n\,\rvert <\varepsilon.$$

Bernard
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    Off topic, but could you give me a hint for why ${\sin n | n \in \mathbb N }$ is dense in $[-1,1]$? –  Aug 17 '15 at 22:12
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    @BolzWeir That follows from the density of the fractional parts ${n\alpha}$ for any fixed irrational $\alpha$. Not only are the values dense, but they follow the same distribution as $\sin X$ where $X$ is a uniform random variable on $[0,2\pi]$. – Erick Wong Aug 17 '15 at 22:17
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    Roughly, it comes from the fact that $\mathbf Z+2\pi\mathbf Z$ is dense in $\mathbf R$, because one proves a subgroup of $\mathbf R$ either is discrete, isomorphicto $\mathbf Z$, or is dense in $\mathbf R$ If $\mathbf Z+2\pi\mathbf Z$ were not dense, it would be discrete, and this would imply $\pi$ is rational. – Bernard Aug 17 '15 at 22:31
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    Even if we didn't know that pi is irrational (a late 18th century discovery) we can show from 3<pi<22/7 that , for infinitely many n, we have n=k.pi +d(n), where k is an integer and |d(n)|<1/7. So |sin n|<sin (1/7) for infinitely many n. – DanielWainfleet Aug 18 '15 at 00:42
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Density (i.e., the irrationality of $\pi$) is not needed.

Let $f\colon\mathbb R$ be any periodic function with $|f(x)\le 1$ for all $x$ and there exists a closed interval $I$ of length $1$ and a number $q<1$ such that $|f(x)|\le q$ for all $x\in I$. Then for any $m\in\mathbb N_0$ $$\lim_{n\to \infty}n^mf(1)f(2)\cdots f(n) = 0 $$

To see this let $p$ be a period of $f$ Then $I, I+1, \ldots , I+\lceil p\rceil-1$ cover a full period of $f$, hence among any $\lceil p\rceil $ consecutive integers $k+i$, $0\le i<\lceil p\rceil$, there is at least one with $|f(k+i)\le q$. As $f(k)|\le 1$ for all other factors, we conclude $$\left|\prod_{k=1}^nf(k)\right| \le q^{\lfloor n/\lceil p\rceil\rfloor}<\frac1q\cdot(\sqrt[\lceil p\rceil]q)^n$$ This exponentially small bound implies the claim.

To apply this to the original problem with $f(x)=\sin x$ observe that one may take for example $I=[-\frac12,\frac12]$ and $q=\sin \frac12<\frac12$.

Hagen von Eitzen
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Hint.

If $$u_n=\sin(1).\sin(2) \dots \sin(n)$$ was having a non vanishing limit $l$ then $$1=\lim\limits_{n \to +\infty} \frac{u_{n+1}}{u_n}= \lim\limits_{n \to +\infty} \sin(n+1)$$ which doesn't make sense as $(\sin(n))_{n \in \mathbb N}$ is dense in $[0,1]$ as you mentionned. So the only potential limit is zero. And indeed the limit is zero again due to the density of $(\sin(n))_{n \in \mathbb N}$ in $[0,1]$.

Regarding the secong sequence $$v_n=n \sin(1).\sin(2) \dots \sin(n)$$ we have with a similar argument that if the limit exists it can only be $0$. Then, again by the density argument, we can find a strictly increasing sequence of integers $(\alpha_n)_{n \in \mathbb N}$ such that $$\vert \sin (\alpha_n) \vert < \frac{1}{2}$$ Hence for $m \ge \alpha_n$ we have $\vert u_m \vert < \frac{1}{2^n}$... Missing some elements to move to the full conclusion...

mathcounterexamples.net
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  • The second sequence does indeed have limit $0$, since the sine product shrinks at least exponentially fast (possibly a little bit faster, even?). – Erick Wong Aug 17 '15 at 22:20
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    What is that "similar argument"? If $s_n$ is a sequence that is dense in $[-1, 1]$, can you really conclude that $ns_1s_2\ldots s_n$ tends to $0$. – Rob Arthan Aug 17 '15 at 22:29
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    To conclude $|u_m| < 1/2^n$, you'll need $m > \alpha_n$, won't you? And the $\alpha_n$ could grow very fast, if $\sin(n)$ were replaced by some other dense sequence contained in $[-1, 1]$. – Rob Arthan Aug 17 '15 at 22:42
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    @Rob Yes I agree and updated the answer... Time to go to bed in France. I'll miss a clean answer for today! – mathcounterexamples.net Aug 17 '15 at 22:45
  • consider x(1)=sin(1) and x(n+1)=(n+1)(sin n)/n. Then |x(n)| is arbitrarily small for infinitely n, so if the limit of second formula exists,it is 0. But |x(n)| may be greater than 1 for some n. – DanielWainfleet Aug 18 '15 at 01:00
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A simple solution is given by considering that: $$ \left|\sin(a-1)\sin(a)\sin(a+1)\right| \leq \frac{5}{17} \tag{1}$$ since: $$ \frac{d}{da}\,\sin(a-1)\sin(a)\sin(a+1) = \frac{1}{2}\cos(a)\left(2+\cos 2-3\cos(2a)\right)\tag{2}$$ so it is not difficult to locate the stationary points of the LHS of $(1)$ and to state: $$ \left| \sin(1)\cdot \sin(2)\cdot\ldots\cdot\sin(n)\right|\leq \left(\frac{2}{3}\right)^{n-2}\tag{3}$$ from which both limits are trivial.

Jack D'Aurizio
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