How to prove $\lim_{n\to \infty} \prod_{i=0}^{n-1} $ $(2+\cos \frac{i \pi}{n})^{\frac{\pi}{n}}$=$\sqrt{3}$ by the sum form of a integration? The integral seems like 
$$
\exp\left(\int_{0}^{\pi}\left[\ln(2+\cos\left(x\right)\right)]\,{\rm d}x\right)
$$
but the above will be a number bigger than $5$ by appraisement.

 A: Taking log, 
\begin{align*}
\log \left( \prod_{i=0}^{n-1}   \left( 2 + \cos \left( \frac{i \pi}{ n } \right ) \right )^{\frac \pi n}\right ) &= 
\sum_{i=0}^{n-1} \frac \pi n  \log \left(  2 + \cos \left( \frac{ i \pi}{n}\right ) \right ) \\ 
 &= \pi  \cdot \frac {1}{n } \sum_{i=0}^{n-1}  \log \left(  2 + \cos \left( \frac{ i \pi}{n}\right ) \right )\\
\end{align*} 
Taking limit 
$$\lim_{n\to\infty } \frac \pi  n \sum_{i=0}^{n-1}  \log \left(  2 + \cos \left( \frac{ i \pi}{n}\right ) \right )  =  \int_0^\pi \log(2 + \cos(x))dx  = \pi  \log \left(\frac{1}{2} \left(2+\sqrt{3}\right)\right) $$
Taking back to $e$, we get 
$$\lim_{n \to \infty } \prod_{k=0}^\infty \left( 2 + \cos \left( \frac{i \pi }{n } \right ) \right )^{\frac \pi n } = e^{\pi  \log \left(\frac{1}{2} \left(2+\sqrt{3}\right)\right) } = \left( \frac 1 2 (  2 + \sqrt 3) \right )^\pi $$
Putting

N[Product[(2 + Cos[k Pi/10^3])^(Pi/10^3), {k, 0, 10^3}]]

on Mathematica produces the value of $7.10988$ which is close to our calculated value so I think what you claim on title must be false.
To evaluate the integral, we proceed in the following way.
$$\int_0^{\pi}  \log \left( 2 + \cos(\theta) \right )d\theta  = \frac 1 2 \int_{-\pi}^{\pi} \log \left( 2 + \cos(\theta) \right )d\theta  = \frac{1}{2} \int_0^{2\pi} \log(2 + \cos(\theta))d\theta $$
From Gauss MVT we get, 
$$\int_0^{2\pi } \left(\log ( \sqrt 3 + 2 + e^{i\theta})+\log ( \sqrt 3 +  2 + e^{-i\theta}) \right ) d\theta  = 4 \pi  \log (\sqrt 3 + 2 )$$
\begin{align*}
 4 \pi  \log (\sqrt 3 + 2 ) &=\int_0^{2\pi } \left(\log (\sqrt 3  + 2 + e^{i\theta})+\log (\sqrt 3 +  2 + e^{-i\theta}) \right ) d\theta \\ 
 &= \int_0^{2\pi} \log \left( ( \sqrt 3+2 )^2 + 1 + ( \sqrt 3 +2)2 \cos \theta  \right )d\theta\\ 
 &= \int_0^{2\pi } \log \left( 4(2 + \sqrt 3) + 2 (2 + \sqrt 3) \cos (\theta) \right )  d\theta \\ 
 &=  \int_0^{2\pi }\log(2(2 + \sqrt 3))d\theta + \int_0^{2\pi } \log(2 + \cos\theta)d\theta\\  
 &= 2 \pi \log (2 (2 + \sqrt 3 )) + \int_0^{2\pi } \log(2 + \cos\theta)d\theta \\  
\end{align*}
Which gives 
$$2\pi \log \left( \frac{(2 + \sqrt 3 )^2}{2 (2 + \sqrt 3 )}\right )= \int_0^{2\pi } \log(2 + \cos\theta)d\theta $$
Taking half of it give the value of our integral as evaluated by Mathematica earlier.

$$\pi \log \left( \frac{2 + \sqrt 3 }{2 }\right )= \int_0^{\pi } \log(2 + \cos\theta)d\theta $$

