Limit of cosines product Let $\displaystyle P_n:=\prod_{k=1}^n \cos\frac{\pi}{k+2}$. Evaluate $\displaystyle\lim_{n\to \infty} P_n$.

I've only shown that the limit is positive.

Let $\vartheta_k:=\pi/(k+2)$. We have $\log P_n=\sum_{k=1}^n \log\cos\vartheta_k$. Now, 
$$\log\cos\vartheta_k=-\frac{1}{2}\log(1+\tan^2\vartheta_k)&gt;-\frac{1}{2}\tan^2\vartheta_k=-\frac{1}{2}\frac{\sin^2\vartheta_k}{\cos^2\vartheta_k}&gt;-2\vartheta_k^2.$$
Therefore we can write that
$$\log P_n&gt;-2\pi^2\sum_{k=1}^n \frac{1}{(k+2)^2}&gt;-2\pi^2\int_2^{\infty}\frac{dx}{x^2}=-\pi^2$$
i.e. $P_n&gt;e^{-\pi^2}$.
We may use the product for $\sin x$ and $\cos x$ to conclude that
\begin{align}
\prod_{n = 1}^\infty \cos\frac{x}{n} &amp;= \prod_{n = 1}^\infty \prod_{k = 0}^\infty \left( 1 - \frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\
&amp;= \prod_{k = 0}^\infty \prod_{n = 1}^\infty \left( 1-\frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\
&amp;= \prod_{k = 0}^\infty \frac{\sin\frac{2x}{2k + 1}}{\frac{2x}{2k + 1}}
\end{align}
which could be used to approximate your product.
We may use the product for $\sin x$ and $\cos x$ to conclude that
\begin{align}
\prod_{n = 1}^\infty \cos\frac{x}{n} &amp;= \prod_{n = 1}^\infty \prod_{k = 0}^\infty \left( 1 - \frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\
&amp;= \prod_{k = 0}^\infty \prod_{n = 1}^\infty \left( 1-\frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\
&amp;= \prod_{k = 0}^\infty \frac{\sin\frac{2x}{2k + 1}}{\frac{2x}{2k + 1}}
\end{align}
which could be used to approximate your product.