Limit related to $ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $? Let $n$ be a positive integer.
Let $b = 2 n - 1$.
Let $x$ be a positive integer.
Define $f(x)$ as :
$$ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $$
Then it appears that
$$ f(b) = \frac{4}{5} + C(n)$$
And $C(n)$ is close to zero.
In fact
$$ \lim_{n \to \infty} C(n) = 0 $$
How do we prove this ??

If you are allowed to use the complex representation of the sine and Pochhammer symbols,
$$f(b) = \prod_{i = 1}^{2n-1} \Big[\sin \left(\frac{\pi  i}{n}\right)+\frac{5}{4}\Big]=-\frac{4}{5} \frac{  \left(\frac{i}{2};e^{\frac{i \pi
   }{n}}\right){}_{2 n} \left(2 i;e^{\frac{i \pi }{n}}\right){}_{2 n}}{4^n\,e^{i \pi  n}}$$which makes
$$C_n=-\frac{4}{5} \Big[1+\frac{  \left(\frac{i}{2};e^{\frac{i \pi
   }{n}}\right){}_{2 n} \left(2 i;e^{\frac{i \pi }{n}}\right){}_{2 n}}{4^n\,e^{i \pi  n}}\Big]$$
For the angles  we know the exact trignometric functions, this generates for $C_n$'s the sequence
$$\left\{\frac{9}{20},-\frac{31}{320},\frac{129}{5120},-\frac{511}{81920},?,-\frac{8191}{20971520},?,-\frac{131071}{5368709120},?,?,?,-\frac{33554431}{351843720888320}\right\}$$ and
$$\log(|C_n|) \sim \frac 12 - b\,n \qquad \text{where} \qquad b=1.38861 \quad \left(\sigma_b=0.00053\right)$$
