Prove-that-lim-n-n-0-t-n-n-e-t-dt-1-2-

Question Number 145399 by Willson last updated on 04/Jul/21

Prove that

\lim_{n \to + \infty} {\int_{0}}^{n} \frac{t^{n}}{n!} e^{- t} dt = \frac{1}{2}

Answered by ArielVyny last updated on 04/Jul/21

\int_{0}^{n} \frac{t^{n}}{n!} e^{- t} d t = \frac{1}{n!} \int_{0}^{+ \infty} e^{- t} t^{n} d t \frac{1}{n!} Γ (n + 1) = \frac{n!}{n!} = 1