How does one show that $\lim_{n \to \infty}\prod_{i=n}^{2n}\left({3\over 2} -y\right)\left({1\over 2}+y\right)=1?$

Question

How does one show that

$$\lim_{n\to \infty}\prod_{i=n}^{2n} \left({3\over 4}+{1\over \pi}\cdot\tan^{-1}{\left(i\over \pi\right)}-{1\over \pi^2}\cdot\tan^{-1}\left({i\over \pi}\right)^2\right)=1\tag1$$

$$\tan^{-1}\left({i\over \pi}\right)^2=\left[\tan^{-1}\left({i\over \pi}\right)\right]^2$$

An attempt:

$$y={1\over \pi}\tan^{-1}\left({i\over \pi}\right)$$

$$\lim_{n\to \infty}\prod_{i=n}^{2n} \left({3\over 4}+y-y^2\right)=1\tag2$$

${3\over 4}+y-y^2=\left({3\over 2} -y\right)\left({1\over 2}+y\right)$

$$\lim_{n \to \infty}\prod_{i=n}^{2n}\left({3\over 2} -y\right)\left({1\over 2}+y\right)=1\tag3$$

$$\lim_{n \to \infty}\prod_{i=n}^{2n}\left({3\over 2} -{1\over \pi}\tan^{-1}\left({i\over \pi}\right)\right)\left({1\over 2}+{1\over \pi}\tan^{-1}\left({i\over \pi}\right)\right)=1\tag4$$ How would we tackle $(4)?$

Answer 1

$$\frac{3}{2}-\frac{1}{\pi}\arctan\frac{k}{\pi} = 1+\frac{1}{\pi}\arctan\frac{\pi}{k},\qquad \frac{1}{2}+\frac{1}{\pi}\arctan\frac{k}{\pi}=1-\frac{1}{\pi}\arctan\frac{\pi}{k} $$ hence you are looking for $$ L=\lim_{n\to+\infty}\prod_{k=n}^{2n}\left(1-\frac{1}{\pi^2}\arctan^2\frac{\pi}{k}\right) $$ but in a neighbourhood of the origin we have $\arctan^2(z)=z^2+O(z^4)$ and $\log(1-z)=-z+O(z^2)$, hence: $$ L = \exp\left[\lim_{n\to +\infty}\sum_{k=n}^{2n}\log\left(1-\frac{1}{\pi^2}\arctan^2\frac{\pi}{k}\right)\right]=\lim_{n\to +\infty}\prod_{k=n}^{2n}\left(1-\frac{1}{k^2}\right)$$ where the last product is a telescopic product, clearly convergent to $1$.