Question Number 143225 by Canebulok last updated on 11/Jun/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{1}}{\left(\mathrm{2021}^{{n}} \right)\left({n}!\right)}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2021}^{{n}} \right)\left(\int_{\mathrm{0}} ^{\:\infty} {t}^{{n}} .{e}^{−{t}} \:\:{dt}\right)} \\ $$
Answered by Dwaipayan Shikari last updated on 11/Jun/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!}={e}^{{x}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2021}\right)^{{n}} {n}!}={e}^{\mathrm{1}/\mathrm{2021}} \\ $$