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Question Number 89092 by 174 last updated on 15/Apr/20

$$Evaluate:\underset{n\to \mathrm{\infty }}{\lim }{e}^{-n}\underset{k=0}{\sum ^n}\frac{n^k}{k!}$$

Commented by abdomathmax last updated on 15/Apr/20

$$wehave{e}^n=\sum _{k=0}^{\mathrm{\infty }}\frac{n^k}{k!}=\sum _{k=0}^n\frac{n^k}{k!}+\sum _{k=n+1}^{\mathrm{\infty }}\frac{n^k}{k!}$$$$=\sum _{k=0}^n\frac{n^k}{k!}+R_n{R}_{n}istherestand{lim}_{n\to +\mathrm{\infty }}Rn=0$$$$\Rightarrow 1=e^{-n}\sum _{k=0}^n\frac{n^k}{k!}+e^{-n}{R}_n\mathrm{\forall }nwepassetolimit$$$$weget{lim}_{n\to +\mathrm{\infty }}{e}^{-n}\sum _{k=0}^n\frac{n^k}{k!}=1$$

Commented by 174 last updated on 15/Apr/20

thanks

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