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 ??
 A: Using
$$
\begin{align}
\sin\left(\frac{k\pi}n\right)+\frac54
&=\frac{e^{ik\pi/n}-e^{-ik\pi/n}}{2i}+\frac54\tag1\\
&=\frac{e^{-ik\pi/n}}{2i}\left(e^{ik\pi/n}+2i\right)\left(e^{ik\pi/n}+i/2\right)\tag2
\end{align}
$$
we get
$$
\begin{align}
\prod_{k=1}^{2n-1}\left(\sin\left(\frac{k\pi}n\right)+\frac54\right)
&=\frac45\prod_{k=0}^{2n-1}\color{#00F}{\frac{e^{-ik\pi/n}}{2i}}\color{#C00}{\left(e^{ik\pi/n}+2i\right)}\color{#090}{\left(e^{ik\pi/n}+i/2\right)}\tag3\\
&=\frac45\color{#00F}{\frac{(-1)^{n-1}}{2^{2n}}}\color{#C00}{\left[z^{2n}-1\right]_{z=2i}}\color{#090}{\left[z^{2n}-1\right]_{z=i/2}}\tag4\\[3pt]
&=\frac45\frac{(-1)^{n-1}}{2^{2n}}\left(2-(-1)^n\left(2^{2n}+2^{-2n}\right)\right)\tag5\\[6pt]
&=\frac45+\color{#90F}{\frac45\left((-1)^{n-1}2^{1-2n}+2^{-4n}\right)}\tag6\\[9pt]
&=\frac45+\color{#90F}{C(n)}\tag7
\end{align}
$$
Explanation:
$(3)$: $\frac45$ of the $k=0$ term is $1$; apply $(2)$
$(4)$: $\prod\limits_{k=0}^{2n-1}\!\left(z+e^{ik\pi/n}\right)=z^{2n}-1$
$(5)$: evaluate at $z=2i$ and $z=i/2$
$(6)$: simplify
$(7)$: the product is $\frac45+C(n)$
Thus,
$$
\begin{align}
C(n)
&=\frac45\left((-1)^{n-1}2^{1-2n}+2^{-4n}\right)\tag8\\[6pt]
&=O\!\left(4^{-n}\right)\tag9
\end{align}
$$
A: 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
$$\log(|C_n|) \sim \frac 12 - b\,n \qquad \text{where} \qquad b=1.38861 \quad \left(\sigma_b=0.00053\right)$$
