Prove that $\lim\limits_{n\to\infty} \prod\limits_{k=1}^n \sin(k)$ converges to $0$

Question

$$\lim_{n\to\infty} \prod_{k=1}^n \sin(k) =0$$ How can I show that this in fact is true? I know that when I take the absolute value of this product than it is strictly decreasing and bounded between 0 and 1, but I want to show that this limit is actually 0.

Answer 1

It suffices to show that there are infinitely many $k \in \mathbb{N}$ such that $|sin(k)| < 1-\varepsilon$ for some fixed $\varepsilon > 0$.

To show that, it suffices to show that if you have a $k_0$ such that $|sin(k_0)| \geq 1 - \varepsilon$ you can always find a $k > k_0$ such that $|sin(k)| < 1 - \varepsilon$. By choosing $\varepsilon$ sufficiently small, this can be seen easily ($k = k_0 + 1$ should actually be sufficient, I guess)

Answer 2

Let's note by $$a_{n}=\left|\prod\limits_{k=1}^n\sin{k}\right|$$ as a result $$a_{n+1}=a_n\cdot\left|\sin{(n+1)}\right|\leq a_n \tag{1}$$ and $$0\leq a_n \leq 1 \tag{2}$$ for $\forall n\in\mathbb{N}$. Exactly as you mentioned, the sequence is monotone and bounded, thus it has a limit.

---

Additionally, $\{\sin{k}\mid k\in\mathbb{N}\}$ is dense in $[-1,1]$, which makes $\{\left|\sin{k}\right|\mid k\in\mathbb{N}\}$ dense in $[0,1]$. This means that for $\forall \varepsilon>0$, there $\exists k(\varepsilon)\in \mathbb{N}$ s.t. $$0<\left|\sin{k(\varepsilon)}\right|<\varepsilon$$ or, using $(2)$ $$0\leq a_{k(\varepsilon)-1}\leq 1 \Rightarrow \\ 0\leq a_{k(\varepsilon)-1}\left|\sin{k(\varepsilon)}\right|\leq \left|\sin{k(\varepsilon)}\right| <\varepsilon \overset{(1)}{\Rightarrow} \\ 0\leq a_{k(\varepsilon)} < \varepsilon$$ and $\forall n \geq k(\varepsilon)$ we have $$0\leq a_{n}\overset{(1)}{\leq} a_{k(\varepsilon)} < \varepsilon$$ Altogether

$\forall \varepsilon>0$, there $\exists k(\varepsilon)\in\mathbb{N}$ s.t. $\forall n\geq k(\varepsilon), n\in\mathbb{N}$ we have $$|a_n - 0| = a_n < \varepsilon$$

But this is the (or one of the) definition of the limit. So $\lim\limits_{n\to\infty}a_n=0$.

Answer 3

Since $$ \mathop {\lim }\limits_{k \to + \infty } \inf \left| {\sin k} \right| = 0 $$ you have that for each $\varepsilon>0$ there exists a sequence $(k_j)_j$ such that $$ \mathop {\lim }\limits_{j \to + \infty } \left| {\sin k_j } \right| = 0 $$ Since, as you noted before, by monotony, we have that $$ \mathop {\lim }\limits_{n \to + \infty } \prod\limits_{k = 1}^n {\left| {\sin k} \right|} = c \geqslant 0 $$ we have that it must be $c=0$. From this it follows that $$ \mathop {\lim }\limits_{n \to + \infty } \prod\limits_{k = 1}^n {\sin k} = 0 $$ Note: the first assertion follows from the density of $(k)_k$, $\mod 2\pi$ due to irrationality of $\pi$.

Answer 4

HINT: Doing some numerical experimentation, it appears that as $n\to \infty$ $$ \left |\prod_{k=1}^n \sin k\right |^{\frac{1}{n}}\to \frac{1}{2} $$ for $n$ large. This follows from the equidistribution theorem, since we have $$\frac{1}{2\pi}\int_0^{2\pi} \log |\sin x|\,dx =\log\frac{1}{2}$$

Answer 5

Group $\sin(2k-1)\sin(2k)=\frac1{2}(\cos(1)-\cos(1-4k))$

You have then

$$\prod\limits_{k=1}^{\infty}\sin(k)=\prod\limits_{k=1}^{\infty}\frac1{2}(\cos(1)-\cos(1-4k))$$

Since $\cos(1)-1\leq\cos(1)-\cos(1-4k)\leq \cos(1)+1$ and

$$\prod\limits_{k=1}^{\infty}\frac{\cos(1)-1}{2}=0$$ $$\prod\limits_{k=1}^{\infty}\frac{\cos(1)+1}{2}=0$$

the limit of $\prod\limits_{k=1}^{\infty}\sin(k)$ by sandwich theorem must be $0$ as well.