Find the limit of $\prod_{k = 4}^{\infty}\cos\left(\pi \over k\right)$ Find the limit of 
$$\prod_{k = 4}^{\infty}\cos\left(\pi \over k\right)$$  
The limit does exist, but I can not get it.
Thanks Willie-Wong &  Lee Mosher for correcting the expression. 
 A: This is somewhat in between a comment and an answer.
To my surprise, the corresponding sum starting at $k = 3$ actually has a name. It is known as the Kepler-Bouwkamp constant or polygon-inscribing constant (OEIS A085365).
Start from a circle of radius $1$. Inscribe a triangle in it and take the in-cirlce.
Inscribe a square in the new circle and then take the in-circle. Repeat this procedure
for every regular $n$-gon, the limiting radius is this Kepler-Bouwkamp constant $K'$.
$$K' \approx 0.11494204485329620070104015765681268475360043148473...$$
The product we have is simply twice of this constant:
$$\prod_{k=4}^\infty \cos\frac{\pi}{k} = 2 K'$$
Aside form this, I can't find any useful information about this sum. The references in the wiki and OEIS link above may have more information for the Kepler-Bouwkamp constant, you should take a look at them.
A: A little addendum to achille hui's answer. We may notice that in a neighbourhood of the origin:
$$ \log(\cos x) = -\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}+\ldots \tag{1} $$
and the complete Taylor series can be recovered from the Weierstrass product:
$$ \cos x = \prod_{k\geq 0}\left(1-\frac{4x^2}{\pi^2(2k+1)^2}\right) \tag{2}$$
leading to:
$$ \begin{eqnarray*}\log(\cos x)&=&-\sum_{k\geq 0}\sum_{n\geq 1}\frac{4^n x^{2n}}{n \pi^{2n} (2k+1)^{2n}}\\&=&-\sum_{n\geq 1}\frac{(4^n-1)\zeta(2n)}{n\pi^{2n}}\,x^{2n},\tag{3}\end{eqnarray*} $$
so, by replacing $x$ with $\frac{\pi}{j}$ and summing over $j\geq 4$ we get:
$$\begin{eqnarray*} \prod_{j\geq 4}\cos\left(\frac{\pi}{j}\right)&=&\exp\left(-\sum_{n\geq 1}(4^n-1)\frac{\zeta(2n)}{n}\left(\zeta(2n)-1-\frac{1}{2^{2n}}-\frac{1}{3^{2n}}\right)\right).\tag{4}\end{eqnarray*}$$
The last sum can be further rearranged. Let $r(m)$ be the arithmetic function that counts the number of ways of writing $m$ as $a\cdot b$ with $a,b\in\mathbb{N}^+$ and $a\geq 4$. We have:
$$\begin{eqnarray*} \sum_{n\geq 1}(4^n-1)\frac{\zeta(2n)}{n}\left(\zeta(2n)-1-\frac{1}{2^{2n}}-\frac{1}{3^{2n}}\right)&=&\sum_{n\geq 1}\frac{4^n-1}{n}\sum_{m\geq 1}\frac{r(m)}{m^{2n}}\\&=&\sum_{m\geq 4}r(m)\log\left(\frac{m^2-1}{m^2-4}\right)\tag{5}\end{eqnarray*}$$
hence:
$$ \prod_{j\geq 4}\cos\left(\frac{\pi}{j}\right) = \prod_{m\geq 4}\left(\frac{m^2-4}{m^2-1}\right)^{r(m)},\tag{6}$$
where $\prod_{m\geq 4}\frac{m^2-4}{m^2-1}$ is a telescopic product that equals $\frac{2}{5}$.
A: Let  $x_n = \prod_{k=4}^{n} \cos(\frac{\pi}{k})$ where $k$ is natural number and $n>3$.
Since $x_n$ is positive for all $n$ and decreasing since $\cos(\frac{\pi}{k})$ is less than $1$, then $x_n$ converges to $\inf \{x_n | n=4,5,...\}$, by monotone convergence theorem.
Now, show that $x_n \rightarrow 0$:
Consider $$x_{n+1} = x_n \cdot \cos(\frac{\pi}{n+1});$$
Since $x_n \rightarrow a$ for some $0 \le a \le \cos(\frac{\pi}{4})$, then $x_{n+1} \rightarrow a $ too, and $\cos(\frac{\pi}{n+1}) \rightarrow 0$.
Therefore $a = \lim \ x_{n+1} = \lim \ x_n  \cdot \lim \ \cos(\frac{\pi}{n+1})  = 0 $.
