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Problem: Find $\lim_{n \to \infty}(\sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n))$.

My idea: I suppose it to be $0$, but how might I go about proving this?

Cameron Williams
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Photon
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    Have a look at this http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1 – mfl Mar 28 '15 at 20:15
  • Note that $\pi$ is irrational and $\sin(x) \in {\pm 1} \Leftrightarrow x = \frac{2k+1}2\pi$. Thus $|\sin k| < 1$ for all $k\in\mathbb N$. Can you conclude? – AlexR Mar 28 '15 at 20:18
  • @Photon do not remove text from a post that is important. Omitting OP's guess for what the answer should be defaces the post. Some users might think that OP didn't even bother to think about this question him/herself which might cause the question to be closed because of your edit. – Cameron Williams Mar 28 '15 at 20:27
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    @AlexR That's actually not enough; you need to know that "enough" of the natural numbers are "far away" from $(2\mathbb{Z}+1)\pi/2$. – Ian Mar 28 '15 at 20:40
  • @Ian Thanks for pointing that out. – AlexR Mar 28 '15 at 20:49

6 Answers6

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$$\sin(k)\sin(k+1)=\frac{1}{2} [\cos(1)-\cos(2k+1)]$$

Now use the fact that $.5 < \cos(1) < .6$ to conclude that

$$\left| \sin(k) \sin(k+1)\right| < 0.8$$ and $$\left| \sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n) \right| \leq 0.8^{\frac{n}{2}}$$

N. S.
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We have $|\sin(n)| \le 1$ for each integer $n \ge 1$.

Consider two consecutive terms $\sin(n),\sin(n+1)$. Their arguments are "$n$ radians" and "$n+1$ radians". When reduced modulo $2\pi$, for at least one of those two arguments, its distances in the circle from $\pi/2$ and $3\pi/2$ both exceed $\frac{1}{2}$ radian. It follows that one of $|\sin(n)|$ or $|\sin(n+1)|$ is less than $$\sin(\frac{\pi-1}{2}) = .877… < .9 $$ Therefore, $$|\sin(n) \sin(n+1)| < .9 $$ and so $$|\sin(1) \cdot \sin(2) \cdot … \cdot \sin(2n+(\text{0 or 1}))| < (.9)^n $$

Lee Mosher
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Since the integers are equidistributed modulo $2\pi$, there are infinitely many $n$ such that $-1/2<\sin(n)<1/2$. Combining this with the fact that $\sin(n)$ is always at most $1$ in absolute value, we find that the limit is $0$.

Equidistribution in fact tells us the exact convergence rate of $|\sin(1)\sin(2)\cdots\sin(n)|$. Taking logarithms, we have that each of the $n$ integers are equally likely to appear from anywhere in the circle, and so $$\sum_{k=1}^{n}\log(|\sin(k)|)=(1+o(1))\frac{n}{2\pi}\int_{0}^{2\pi}\log(|\sin(x)|)dx$$ $$=-n\log2+o(n),$$ and so $$\biggr|\prod_{k=1}^n \sin(k)\biggr|=\left(\frac{1}{2}\right)^{n(1+o(1))}.$$

Eric Naslund
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Hint. Let $n_k = \lfloor \pi k \rfloor$ so that $(n_k)$ is strictly increasing and $|n_k - \pi k| \leq 1$. Can you check that

$$ |\sin n_k| \leq \sin 1 < 1 ?$$

How this leads to the conclusion?

Sangchul Lee
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Consider that in every interval of the form $[\pi n-\frac{1}2,\pi n+\frac{1}2]$ there is at least one integer. Call the set of integers within such an interval $S$ and note that, as there are infinitely many such (disjoint) intervals, $S$ is infinite. However, in such intervals $|\sin(x)|<\sin(\frac{1}2)<1$. If we let $r=\sin(\frac{1}2)$, this, together with the fact that $|\sin(x)|\leq 1$, it follows that a product including the sines of $k$ members of $S$ has absolute value at most $r^k$. Since $k$ goes to infinity, the products can be constrained to any neighborhood of $0$, hence the product converges to $0$.

One can easily make this argument as rigorous as they desire.

Milo Brandt
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First off observe that $|\sin n| \leq 1$ for all $n$. Then can you show that there are infinitely many $n$ which are a bounded distance from 1 (i.e. at most equal to $1 - \epsilon$ for some $\epsilon > 0$)?

Then it follows that $$\prod_{n \in \mathbb N} |\sin n \,| \leq \prod_{n \in \mathbb{N}} (1 - \epsilon) = \lim_{n \rightarrow \infty} (1-\epsilon)^n$$

dalastboss
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