hawatajwizp
2022-06-26
Answered

Show that $\underset{n\to \mathrm{\infty}}{lim}\prod _{i=n}^{Bn}\frac{\mathrm{arctan}(i\varphi )}{\mathrm{arccos}\left(\frac{\varphi}{i}\right)}={B}^{\frac{2}{\pi}}$

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britspears523jp

Answered 2022-06-27
Author has **28** answers

Denote $f(x)=\mathrm{log}\frac{\mathrm{arctan}(x\varphi )}{\mathrm{arccos}\left(\frac{\varphi}{x}\right)}$. Expanding into series

$f(x)=\frac{2}{\pi x}+O({x}^{-2}),\phantom{\rule{1em}{0ex}}x\to \mathrm{\infty},$

and using the Euler's approximation for the harmonic sums

$\sum _{i=1}^{n}\frac{1}{i}=\gamma +\mathrm{log}n+O(1/n),$

where $\gamma $ is the Euler's constant, gives

$\mathrm{log}\underset{n\to \mathrm{\infty}}{lim}\prod _{i=n}^{Bn}\frac{\mathrm{arctan}(i\varphi )}{\mathrm{arccos}\left(\frac{\varphi}{i}\right)}=\underset{n\to \mathrm{\infty}}{lim}\sum _{i=n}^{Bn}f(i)=\underset{n\to \mathrm{\infty}}{lim}[\frac{2}{\pi}(\sum _{i=1}^{Bn}\frac{1}{i}-\sum _{i=1}^{n-1}\frac{1}{i})+O({n}^{-1})]=$

$=\underset{n\to \mathrm{\infty}}{lim}[\frac{2}{\pi}(\mathrm{log}Bn-\mathrm{log}n)+O({n}^{-1})]=\frac{2}{\pi}\mathrm{log}B.$

$f(x)=\frac{2}{\pi x}+O({x}^{-2}),\phantom{\rule{1em}{0ex}}x\to \mathrm{\infty},$

and using the Euler's approximation for the harmonic sums

$\sum _{i=1}^{n}\frac{1}{i}=\gamma +\mathrm{log}n+O(1/n),$

where $\gamma $ is the Euler's constant, gives

$\mathrm{log}\underset{n\to \mathrm{\infty}}{lim}\prod _{i=n}^{Bn}\frac{\mathrm{arctan}(i\varphi )}{\mathrm{arccos}\left(\frac{\varphi}{i}\right)}=\underset{n\to \mathrm{\infty}}{lim}\sum _{i=n}^{Bn}f(i)=\underset{n\to \mathrm{\infty}}{lim}[\frac{2}{\pi}(\sum _{i=1}^{Bn}\frac{1}{i}-\sum _{i=1}^{n-1}\frac{1}{i})+O({n}^{-1})]=$

$=\underset{n\to \mathrm{\infty}}{lim}[\frac{2}{\pi}(\mathrm{log}Bn-\mathrm{log}n)+O({n}^{-1})]=\frac{2}{\pi}\mathrm{log}B.$

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