Question Number 166013 by mathlove last updated on 11/Feb/22
Answered by JDamian last updated on 11/Feb/22
$${x}_{{r}} ={e}^{{i}\frac{\pi}{\mathrm{2}^{{r}} }} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} \centerdot\centerdot\centerdot{x}_{\infty} =\:{e}^{{i}\pi\underset{{r}} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{{r}} }} =\:{e}^{{i}\pi\centerdot\mathrm{1}} \:=\:−\mathrm{1} \\ $$