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Question Number 143225 by Canebulok last updated on 11/Jun/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left({2021}^n\right)(n!)}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left({2021}^n\right)({\int }_0^{\mathrm{\infty }}{t}^n.e^{-t}dt)}$$

Answered by Dwaipayan Shikari last updated on 11/Jun/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}=e^x$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(2021\right)}^{n}n!}=e^{1/2021}$$

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