Arianna Villegas

2022-03-29

$\frac{1}{4}\mathrm{tan}\left(\frac{\pi }{8}\right)+\frac{1}{8}\mathrm{tan}\left(\frac{\pi }{16}\right)+\frac{1}{16}\mathrm{tan}\left(\frac{\pi }{32}\right)+\cdots \cdots \mathrm{\infty }$
Try:
$\mathrm{cos}x\cdot \mathrm{cos}\left(\frac{x}{2}\right)\cdot \mathrm{cos}\left(\frac{x}{{2}^{2}}\right)\cdots \cdots \mathrm{cos}\left(\frac{x}{{2}^{n-1}}\right)=\frac{1}{{2}^{n-1}}\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(\frac{x}{{2}^{n-1}}\right)}$
Could some help me how to solve ahead

Harry Gibson

Using the identity:
$\mathrm{cot}x-\mathrm{tan}x=2\mathrm{cot}2x⇒\mathrm{tan}x=\mathrm{cot}x-2\mathrm{cot}2x$
one obtains a telescoping sum:
$S\left(x\right)\phantom{\rule{0.222em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{tan}\left(\frac{x}{{2}^{n}}\right)}{{2}^{n}}=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{\mathrm{cot}\left(\frac{x}{{2}^{n}}\right)}{{2}^{n}}-\frac{\mathrm{cot}\left(\frac{x}{{2}^{n-1}}\right)}{{2}^{n-1}}\right)$
$=-2\mathrm{cot}\left(2x\right)+\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{cot}\left(\frac{x}{{2}^{n}}\right)}{{2}^{n}}=\frac{1}{x}-2\mathrm{cot}\left(2x\right)$
Applying this to your case one obtains
$\frac{1}{4}S\left(\frac{\pi }{8}\right)=\frac{2}{\pi }-\frac{1}{2}$

Do you have a similar question?