The limit of a sequence $\lim_{n\rightarrow \infty}\prod_{k=0}^{n-1}( 2+\cos \frac{k\pi }{n})^{\pi/n}$. $$
\lim_{n \to \infty}\prod_{k = 0}^{n - 1}
\left[\,2 + \cos\left(k\pi \over n\right)\right]^{\pi/n}\ =\ ?
$$
Please give some hints.
 A: $$y=\lim_{n\rightarrow \infty} \prod_{k=1}^{n-1}\left(2+\cos\frac{k\pi}{n}\right)^{\pi/n} \Rightarrow \ln y=\lim_{n\rightarrow \infty} \frac{\pi}{n}\sum_{k=1}^{n-1} \ln\left(2+\cos\frac{k\pi}{n}\right)$$
$$\Rightarrow \ln y=\pi \int_0^{\pi}\ln(2+\cos x)\,dx$$
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
$$\boxed{y=\left(\dfrac{2+\sqrt{3}}{2}\right)^{\pi}}$$
A: Evaluate the logarithm of your expression as the limiting value of a Riemann sum, factor out the $2$ inside the logarithm appearing in the integrand, then expand the logarithm function into its Maclaurin series and finally integrate the series term by term.
$$\int_{0}^{\pi}\ln(2+\cos(x)) dx=\int_{0}^\pi\ln(2)+\ln(1+\frac{\cos(x)}{2})dx$$
$$=\pi\ln(2)+\int_{0}^\pi\ln(1+\frac{\cos(x)}{2}) dx=\pi\ln(2)+\int_{0}^\pi\sum_{n=1}^\infty\frac{(-1)^{n-1}\cos(x)^n}{n2^n}dx$$
$$=\pi\ln(2)+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n2^n}\int_{0}^\pi\cos(x)^ndx$$
$$=\pi\ln(2)+\sum_{n=1}^\infty\frac{(-1)^{2n-1}}{2n2^{2n}}\int_{0}^\pi\cos(x)^{2n}dx$$
$$\text{Since the integrals vanish at odd terms.}$$
$$=\pi\ln(2)-\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n4^n}\int_{0}^\pi\cos(x)^{2n} dx$$
$$=\pi\ln(2)-\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n4^n}\frac{\binom{2n}{n}\pi}{4^n}=\pi\ln(2)-\frac{\pi}{2}\sum_{n=1}^\infty\frac{\binom{2n}{n}}{n16^n}$$
$$=\pi\ln(2)-\frac{\pi}{2}(4\ln(2)-2\ln(2+\sqrt{3}))=\pi\ln(2)-2\pi\ln(2)+\pi\ln(2+\sqrt{3}))$$
$$=-\pi\ln(2)+\pi\ln(2+\sqrt{3})=\pi(\ln(2+\sqrt{3})-\ln(2))=\pi\ln(1+\frac{\sqrt{3}}{2})$$
Thus we have that:
$$\lim_{n\rightarrow \infty}\prod_{k=0}^{n-1}\left( 2+\cos \frac{k\pi }{n}\right)^{{\pi}/{n}}=(1+\frac{\sqrt{3}}{2})^{\pi}$$
A: For any $a \in (1,\infty)$, consider the equation
$$\frac{x+x^{-1}}{2} = a\quad\iff\quad x^2 - 2ax + 1 = 0$$
It has two roots and one of it, $x = a + \sqrt{a^2-1}$, is outside the unit circle.
Notice
$$z^{2n}-1 
= (z^2 - 1)\prod_{k=1}^{n-1}\left(z-e^{i\frac{k\pi}{n}}\right)\left(z-e^{-i\frac{k\pi}{n}}\right)
= (z^2 - 1)z^{n-1}\prod_{k=1}^{n-1}\left(z + \frac{1}{z} - 2\cos\frac{k\pi}{n}\right)
$$
Substitute $z$ by $-x = -\Big(a + \sqrt{a^2-1}\Big)$, this leads to
$$
\prod_{k=1}^{n-1}\left(a + \cos\frac{k\pi}{n}\right)
= \prod_{k=1}^{n-1}\left(\frac{x+x^{-1}}{2} + \cos\frac{k\pi}{n}\right)
=  \frac{x^{2n}-1}{(x^2 - 1)(2x)^{n-1}}
$$
For large $n$, the numerator is dominated by the $x^{2n}$ term. As a result,
$$
\lim_{n\to\infty}\prod_{k=0}^{n-1}\left(a + \cos\frac{k\pi}{n}\right)^{\pi/n}
= \lim_{n\to\infty}\left(\frac{(x+1)(x^{2n}-1)}{(x-1)(2x)^n}\right)^{\pi/n}
= \left(\frac{x}{2}\right)^{\pi}
= \left(\frac{a + \sqrt{a^2-1}}{2}\right)^{\pi}
$$
Substitute $a$ by $2$, the desired limit is
$\displaystyle\;\left(\frac{2+\sqrt{3}}{2}\right)^\pi\approx
7.09761749580862646182\ldots
.$
