Infinite product $\prod\limits_{n=1}^\infty\cos( 1/n)$

Question

Infinite product $\prod\limits_{n=1}^\infty\cos( 1/n)$

My attempt is to use the complex representation and the Maclaurin Series of $\cos(1/n)$ but I could not find the formula $$\prod_{n=0}^\infty (1 + {b_n})$$ Any hints please.

Answer 1

$\prod_{n=1} ^{\infty } (1+b_n)$ converges if $\sum_{n=1} ^{\infty } |b_n| <\infty$. If $b_n=\cos (\frac 1 n) -1$ then $\sum_{n=1} ^{\infty } |b_n| \leq \sum_{n=1} ^{\infty }\frac 1 {2n^{2}} <\infty$ so the product converges. [I have used the inequality $0\leq 1-\cos x \leq \frac {x^{2}} 2$].

Answer 2

You could also use this inequality, $\forall x \in \left [0, \frac{\pi}{2} \right ]$: $$0\leq\cos(x)\leq e^{-\frac{x^2}{2}}$$ since $0<\frac{1}{n}<\frac{\pi}{2}, \forall n\in \mathbb{N}$. As a result $$0\leq\prod\limits_{n=1}^\infty\cos\left(\frac{1}{n}\right)= \lim\limits_{k\rightarrow\infty}\prod\limits_{n=1}^k\cos\left(\frac{1}{n}\right)\leq \lim\limits_{k\rightarrow\infty}\prod\limits_{n=1}^k e^{-\frac{1}{2n^2}}=\\ \lim\limits_{k\rightarrow\infty}e^{-\frac{1}{2}\left(\sum\limits_{n=1}^k\frac{1}{n^2}\right)}=\frac{1}{e^{\frac{1}{2}\left(\sum\limits_{n=1}^\infty\frac{1}{n^2}\right)}}<\infty$$ since $0<\sum\limits_{n=1}^\infty\frac{1}{n^2}<\infty$.

Answer 3

$\sum_{n\geq 1}\log\cos\frac{1}{n}$ is convergent since $\log\cos\frac{1}{n}\sim -\frac{1}{2n^2}$ as $n\to +\infty$, hence your product is convergent.
Since $$ \cos(x)=\prod_{a\geq 0}\left(1-\frac{4x^2}{(2a+1)^2 \pi^2}\right) $$ we have $$ \log\cos\tfrac{1}{n} = -\sum_{a\geq 0}\sum_{b\geq 1}\frac{4^b}{b(2a+1)^{2b} \pi^{2b}n^{2b}}=-\sum_{b\geq 1}\frac{(4^b-1)\zeta(2b)}{b\pi^{2b}n^{2b}} $$ and $$ \prod_{n\geq 1}\cos\tfrac{1}{n} = \exp\left[-\sum_{b\geq 1}\frac{(4^b-1)\zeta(2b)^2}{b \pi^{2b}}\right]>\left(\frac{\pi^2-4}{\pi^2-1}\right)^{\pi^4/36}>\frac{17}{52}. $$ In a more elementary fashion $$ \prod_{n\geq 1}\cos\tfrac{1}{n}> \prod_{n\geq 1}\left(1-\frac{1}{2n^2}\right) = \frac{\sqrt{2}}{\pi}\sin\frac{\pi}{\sqrt{2}}>\frac{24}{67},$$ $$ \prod_{n\geq 1}\cos\tfrac{1}{n}< \prod_{n\geq 1}\left(1-\frac{1}{4n^2}\right)^2 = \frac{4}{\pi^2}<\frac{28}{67}.$$