# Does $\prod_{k=3}^{\infty}\cos\frac{2\pi}{n!}$ have a closed-form solution?

How to compute this please? $$\prod_{k=3}^{\infty}\cos\frac{2\pi}{k!}$$

I just know it is convergent.

• @Dahaka Convergence is obvious. – Szeto Oct 29 '18 at 12:16
• The product of real parts is not the real part of the product. – Sam Streeter Oct 29 '18 at 12:38
• @Dahaka, Sam Street's is correct here, and it invalidates your argument. However, your argument can be salvaged; it just needs a double sum. – Paul Oct 29 '18 at 12:44
• @Dahaka For convergence, notice $\displaystyle\;\cos\frac{2\pi}{k!} = 1 + O\left(\frac{1}{(k!)^2}\right)$ and $$\sum\limits_{k=1}^\infty |a_k| < \infty \iff \prod\limits_{k=1}^\infty (1+a_k) \text{ converges absolutely }\implies \prod\limits_{k=1}^\infty (1+a_k)\text{ converges }$$ – achille hui Oct 29 '18 at 12:52
• @achillehui also $\cos(2\pi/k!)\in (0,1)$ for $k\ge 3$ so $$\lim_{m\to\infty}\left|\prod_{k=3}^m\cos(2\pi/k!)\right|\le 1$$ – Masacroso Oct 29 '18 at 13:43