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 &amp;  Lee Mosher for correcting the expression.
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.
Let  $x_n = \prod_{k=4}^{n} \cos(\frac{\pi}{k})$ where $k$ is natural number and $n&gt;3$.

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 $.