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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, $$\log\cos\vartheta_k=-\frac{1}{2}\log(1+\tan^2\vartheta_k)>-\frac{1}{2}\tan^2\vartheta_k=-\frac{1}{2}\frac{\sin^2\vartheta_k}{\cos^2\vartheta_k}>-2\vartheta_k^2.$$ Therefore we can write that $$\log P_n>-2\pi^2\sum_{k=1}^n \frac{1}{(k+2)^2}>-2\pi^2\int_2^{\infty}\frac{dx}{x^2}=-\pi^2$$ i.e. $P_n>e^{-\pi^2}$.

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Your second inequality is not right. I think you need to sum $1/(k+2)^2$. – Ron Gordon Jul 28 '13 at 15:12
Typo, you're right. – user72870 Jul 28 '13 at 15:13
What is the source of this problem? Do you know if a "closed form" exists? – Aryabhata Jul 28 '13 at 15:56
A friend of mine gave me the problem... – user72870 Jul 28 '13 at 16:00
possibly useful fact: $\int \tan(x) dx = -\log(\cos(x))+C$ – Omnomnomnom Jul 28 '13 at 16:39

We may use the product for $\sin x$ and $\cos x$ to conclude that \begin{align} \prod_{n = 1}^\infty \cos\frac{x}{n} &= \prod_{n = 1}^\infty \prod_{k = 0}^\infty \left( 1 - \frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\ &= \prod_{k = 0}^\infty \prod_{n = 1}^\infty \left( 1-\frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\ &= \prod_{k = 0}^\infty \frac{\sin\frac{2x}{2k + 1}}{\frac{2x}{2k + 1}} \end{align} which could be used to approximate your product.

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