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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 $(4)?$

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$$\frac{3}{2}-\frac{1}{\pi}\arctan\frac{k}{\pi} = 1+\frac{1}{\pi}\arctan\frac{\pi}{k},\qquad \frac{1}{2}+\frac{1}{\pi}\arctan\frac{k}{\pi}=1-\frac{1}{\pi}\arctan\frac{\pi}{k} $$ 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 $1$.

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    I feel like half of your recent answers have involved $\arctan^2(x)$, or at least some series for $\arctan(x) $. +1 to you sir. – 2017-02-13
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    Applying $\arctan^2(x)$ and logarithm series we get $$L =\lim_{n\to +\infty}\exp\left[-\sum_{k=n}^{2n}\left(\frac{1}{k^2}\right)\right]$$ $$L =\lim_{n\to +\infty}\prod_{k=n}^\infty \exp\left(-k^{-2}\right)$$ Did you simply apply the series $\exp(x) = 1+x+O(x^2)$ here? Moreover, how are we rigorously justified in ignoring all higher order terms of the three series expansions? – 2017-02-13
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    @BrevanEllefsen: I am tacitly exploiting the fact that $$\sum_{k\geq n}\frac{1}{k^2} = O\left(\frac{1}{n}\right),$$ hence we may neglect high-order terms. – 2017-02-13