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Show that $\underset{n\to \mathrm{\infty }}{lim}\prod _{i=n}^{Bn}\frac{\mathrm{arctan}\left(i\varphi \right)}{\mathrm{arccos}\left(\frac{\varphi }{i}\right)}={B}^{\frac{2}{\pi }}$
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britspears523jp
Denote $f\left(x\right)=\mathrm{log}\frac{\mathrm{arctan}\left(x\varphi \right)}{\mathrm{arccos}\left(\frac{\varphi }{x}\right)}$. Expanding into series
$f\left(x\right)=\frac{2}{\pi x}+O\left({x}^{-2}\right),\phantom{\rule{1em}{0ex}}x\to \mathrm{\infty },$
and using the Euler's approximation for the harmonic sums
$\sum _{i=1}^{n}\frac{1}{i}=\gamma +\mathrm{log}n+O\left(1/n\right),$
where $\gamma$ is the Euler's constant, gives
$\mathrm{log}\underset{n\to \mathrm{\infty }}{lim}\prod _{i=n}^{Bn}\frac{\mathrm{arctan}\left(i\varphi \right)}{\mathrm{arccos}\left(\frac{\varphi }{i}\right)}=\underset{n\to \mathrm{\infty }}{lim}\sum _{i=n}^{Bn}f\left(i\right)=\underset{n\to \mathrm{\infty }}{lim}\left[\frac{2}{\pi }\left(\sum _{i=1}^{Bn}\frac{1}{i}-\sum _{i=1}^{n-1}\frac{1}{i}\right)+O\left({n}^{-1}\right)\right]=$
$=\underset{n\to \mathrm{\infty }}{lim}\left[\frac{2}{\pi }\left(\mathrm{log}Bn-\mathrm{log}n\right)+O\left({n}^{-1}\right)\right]=\frac{2}{\pi }\mathrm{log}B.$