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Question Number 160136 by alcohol last updated on 25/Nov/21

$$1+4+\frac{16}{2}+\frac{64}{6}+...+\frac{4^n}{n!}=?$$

Answered by puissant last updated on 25/Nov/21

$$S_n=1+4+\frac{16}{2}+\frac{64}{6}+....+\frac{4^n}{n!}=\underset{k=0}{\sum ^n}\frac{4^k}{k!}$$$$\underset{n\to +\mathrm{\infty }}{\lim }{S}_n=\underset{n\to +\mathrm{\infty }}{\lim }\underset{k=0}{\sum ^n}\frac{4^k}{k!}=e^4...$$

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