TITLE: The limit of a sequence $\lim_{n\rightarrow \infty}\prod_{k=0}^{n-1}( 2+\cos \frac{k\pi }{n})^{\pi/n}$.
QUESTION [9 upvotes]: $$
\lim_{n \to \infty}\prod_{k = 0}^{n - 1}
\left[\,2 + \cos\left(k\pi \over n\right)\right]^{\pi/n}\ =\ ?
$$
Please give some hints.

REPLY [7 votes]: $$y=\lim_{n\rightarrow \infty} \prod_{k=1}^{n-1}\left(2+\cos\frac{k\pi}{n}\right)^{\pi/n} \Rightarrow \ln y=\lim_{n\rightarrow \infty} \frac{\pi}{n}\sum_{k=1}^{n-1} \ln\left(2+\cos\frac{k\pi}{n}\right)$$
$$\Rightarrow \ln y=\pi \int_0^{\pi}\ln(2+\cos x)\,dx$$
Let 
$\displaystyle I=\int_0^{\pi} \ln(2+\cos x)\,dx=\int_0^{\pi} \ln(2-\cos x)\,dx$
$\displaystyle \Rightarrow 2I=\int_0^{\pi} \ln(4-\cos^2x)\,dx \Rightarrow I=\frac{1}{2}\int_0^{\pi}\ln(4-\cos^2x)\,dx=\int_0^{\pi/2} \ln(4-\cos^2x)\,dx$
Consider,
$$J(a)=\int_0^{\pi/2}\ln(a-\cos^2x)\,dx $$
$$\displaystyle \begin{aligned}
\Rightarrow J'(a)&=\int_0^{\pi/2} \frac{dx}{a-\cos^2x} \\
&=\int_0^{\pi/2} \frac{\sec^2x}{a-1+a\tan^2x}\,dx \\
&=\int_0^{\infty} \frac{dt}{a-1+at^2}\,\,\,(t=\tan x)\\
&=\frac{\pi}{2}\frac{1}{\sqrt{a}\sqrt{a-1}}
\end{aligned}$$
$$\displaystyle \Rightarrow J(a)=\pi\ln(\sqrt{a-1}+\sqrt{a})+C$$
With $J(1)=-\pi\ln 2$, $C=-\pi\ln2$, hence,
$$J(a)=\pi\ln\left(\frac{\sqrt{a-1}+\sqrt{a}}{2}\right)$$
$$\Rightarrow I=J(4)=\pi\ln\left(\frac{2+\sqrt{3}}{2}\right)$$
i.e
$$\boxed{y=\left(\dfrac{2+\sqrt{3}}{2}\right)^{\pi}}$$