Question Number 98286 by M±th+et+s last updated on 12/Jun/20
$${justify}: \\ $$$$\underset{{x}\rightarrow+\infty} {{lim}}\:\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left({k}−\mathrm{1}\right)!\:{sin}\left({x}−\frac{\pi}{\mathrm{2}}{k}\right)}{{x}^{{k}} }=\frac{\pi}{\mathrm{2}} \\ $$