$\lim_{n\to\infty}\{\frac{3}{n}\sum_{k=1}^n[1+8\sin^2(\frac{k\pi}{n})]^{-1}\}^{2^n}$ How to calculate the limit below?
$$\lim_{n\to\infty}\{\frac{3}{n}\sum_{k=1}^n[1+8\sin^2(\frac{k\pi}{n})]^{-1}\}^{2^n}$$
Since I used Riemann integration to work out that $$\lim_{n\to\infty}[\frac{3}{n}\sum_{k=1}^n(1+8\sin^2(\frac{k\pi}{n}))^{-1}]=1$$, I've been trying to express $$\frac{3}{n}\sum_{k=1}^n(1+8\sin^2(\frac{k\pi}{n}))^{-1}=(1+\frac{C}{2^n}+o(\frac{1}{2^n})),\quad n\to+\infty$$
Can anyone render me some hints?
 A: Now to prove Claude's observation and stated explicitly by the proposer in a comment.
Start with the known (Hansen, Table of Series and Products, eq. 91.1.17)
$$ \prod_{k=0}^{n-1} \sinh^2{y} + \sin^2(x+k\,\pi/n) = 2^{1-2n}\big(\cosh{(2ny)} - \cos{(2nx)} \big)$$
Set x=0, note that sum can start with 1 and end with n, and take the logarithmic derivative,
$$ 2\, \cosh{y}\,\sinh{y} \, \sum_{k=1}^n \frac{1}{\sinh^2{y}  + \sin^2(k\,\pi/n)}
= 2n \frac{\sinh{2ny}}{\cosh{(2ny)} - 1} $$
Solving $1/\sinh^2{y} = 8 \implies y=\log{2}/2.$  Algebra and hyperbolic trig ID  completes the proof of
$$ \sum_{k=1}^n \frac{1}{1 + 8\sin^2(k\,\pi/n)}=\frac{n}{3}\,\coth{\big(\frac{n}{2}\log{2}\big)}=\frac{n}{3} \frac{2^n+1}{2^n-1} $$
A: From what it seems$$\frac{3 }{n}\sum _{k=1}^n \frac{1}{1+8 \sin ^2\left(\frac{\pi  k}{n}\right)}=\frac{2^n+1}{2^n-1}=1+\frac 2 {2^n}+ \cdots$$
So, now we consider
$$S=\left(\frac{2^n+1}{2^n-1}\right)^{2^n}$$ Let $x=2^n$ making
$$S=\left(\frac{x+1}{x-1}\right)^{x}\implies \log(S)=x \log\left(\frac{x+1}{x-1}\right)=x \log\left(1+\frac{2}{x-1}\right)$$ So, by Taylor
$$\log(S)=2+\frac{2}{3 x^2}+\frac{2}{5 x^4}+O\left(\frac{1}{x^6}\right)$$
$$S=e^{\log(S)}=e^2+\frac{2 e^2}{3 x^2}+\frac{28 e^2}{45
   x^4}+O\left(\frac{1}{x^6}\right)\qquad \text{with}\qquad x=2^n$$
Trying for $n=4$, the exact result is
$$\frac{48661191875666868481}{6568408355712890625}\approx 7.40836885$$ while the above truncated formula gives
$$\frac{739207 }{737280}e^2 \approx 7.40836859$$
