The infinite product of the sines of all positive integers is zero

Question

I tried to prove that $\prod\limits_{k = 1}^\infty {\sin k = 0}$

Define $$P(m)=\prod\limits_{k = 1}^m {\left|\sin k\right|}$$ then I take the logarithm of both sides $$\log P(m)=\log \prod\limits_{k = 1}^m {\left|\sin k\right|}$$ which can be written as $$a_m=\log P(m)=\sum\limits_{k = 1}^m {\log\,\left|\sin k\right|}$$ for any $k\in\mathbb{N}$ we have $0<\left|\sin k\right|<1$, so all terms of the sum above is strictly negative.

I say that the sequence diverges: $\left\{a_m\right\}_{m\in\mathbb{N}}\to\,-\infty$

This is a delicate point on which I have some doubts.

I know that $$\int\limits_0^\pi {\log \left|\sin x\right|{\text{ d}}x} = - \pi \log 2$$ and that the function $\log \left|\sin x\right|$ is periodic with period $\pi$, so the integral from 1 to $\infty$ diverges to $- \infty$. Then the series diverges as well for the integral test for convergence.

As $\log P(m)\to\, -\infty$ we can say that $P(m)\to\,0$ as $m\to\,\infty$

Therefore $$\mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\left| {\sin k} \right| = \mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\sin k = \prod\limits_{k = 1}^\infty {\sin k = } 0} }.\quad \blacksquare$$

Answer 1

One can prove using irrationality of $\pi$ that there are infinitely many natural numbers $\{n_k,k\geqslant 1\}$ such that $|\sin(n_k)|<1/2$. Then $$\left|\prod_{l=1}^{n_k}\sin l\right|\leqslant \left|\prod_{j=1}^k\sin n_j\right|\leqslant \frac 1{2^k}.$$ Conclude taking the limit $k\to +\infty$.

Answer 2

Since $$|\sin(2k-1)\sin(2k)| = \left|\frac{\cos 1-\cos(4k-1)}{2}\right| \le \frac{1+\cos 1}{2} < 1$$ we have $$\begin{align} & 0 \le \left|\; \prod_{k=1}^{n} \sin k\;\right| \le \left|\; \prod_{k=1}^{\lfloor\frac{n}{2}\rfloor} \sin(2k-1)\sin(2k)\;\right| \le \left(\frac{1+\cos 1}{2}\right)^{\lfloor\frac{n}{2}\rfloor} \\ \implies & 0 \le \liminf_{n\to\infty} \left|\; \prod_{k=1}^{n} \sin k\;\right| \le \limsup_{n\to\infty} \left|\; \prod_{k=1}^{n} \sin k\;\right| \le \lim_{n\to\infty} \left(\frac{1+\cos 1}{2}\right)^{\lfloor\frac{n}{2}\rfloor} = 0\\ \implies &\lim_{n\to\infty} \prod_{k=1}^{n} \sin k = 0 \end{align}$$

Answer 3

To complete your idea which is almost correct: note that the integral test for convergence is applicable only for monotone decreasing function so in our example you can't use this test, however we know that the sequence $(\log|\sin k|)$ does not converge to $0$ since the sequence $(\sin k)$ has not even a limit so the series $$\sum_{k=1}^\infty \log|\sin k|$$ is divergent, moreover its general term $\log|\sin k|$ has a negative sign hence we have $$\sum_{k=1}^\infty \log|\sin k|=-\infty$$ which allows us to conclude

Answer 4

We have $\forall n\in \Bbb N :0<\left|\sin(n)\right|<1$. Therefore $P(m)$ is nonnegative and decreasing. Thus it converges.

Now, show that $0$ is a limit point of $\left|\sin(\Bbb N)\right|$. If you already know that $\sin(\Bbb N)$ is dense in $[-1,1]$, you can use that. Otherwise, check one of the many proofs on this site. The first step is usually to show that $0$ is a limit point.

It now follows that no matter what $\epsilon > 0$ we choose, the limit of $P(m)$ is smaller. Conclude that the limit is $0$.

Answer 5

$$\prod_{k=1}^n \left \vert \sin(k) \right \vert = \left(\prod_{k=1}^n \left(1-\cos^2(k)\right) \right)^{1/2}$$ And since $\displaystyle \sum_{k=1}^n \cos^2(k)$ diverges we can conclude from here, that the sequence converges to $0$.

Answer 6

Let $$x_n=\prod_{k=1}^{n}|\sin(k)|,$$ then the sequences $\{x_n\}$ is decreasing and $x_n>0$. So by the monotone bounded convergence theorem, the limit $a:=\lim_{n\to}x_n$ exists. We claim that $a=0$, otherwise, if $a\neq0$, then $$\lim_{n\to\infty}|\sin(n)|=\lim_{n\to\infty}\frac{x_n}{x_{n-1}}=1.$$ This implies $\lim_{n\to}\cos(n)=0$ due to $\sin^2(n)+\cos^2(n)=1$. So we get $$1=\lim_{n\to\infty}|\sin(2n)|=2\lim_{n\to\infty}|\sin(n)\cos(n)|=0,$$ which is a contradiction. So $a=0$.

Answer 7

Actually if you find a real number a with $0<a<1$ and any strictly increasing sub-sequence of integers, say $p(k)$ with the property that $|\sin p(k)|<a,k=1,2,\cdots$, $$b_m=\prod_{k=1}^m|\sin p(k)|<a^m\to 0$$ You are done because you can decompose the natural numbers $N$ as $A \bigcap (N-A)$ where $A=\{ p(1),p(2),\cdots k=1,2,...\}$ and you have the product over A converging to zero and the product over $N-A$ bounded by 1.

Q.E.D.