<s>[INST] 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.
 [/INST] 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
$\displaystyle\;\left(\frac{2+\sqrt{3}}{2}\right)^\pi\approx
7.09761749580862646182\ldots
.$
</s>