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How to calculate $\lim_{n\to\infty} \frac{2^n}{n!}$? I tried using $\lim \frac{x^n}{a^x} = 0$ but it didn't work

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The limit is $0$, since factorial growth is faster than exponential growth.

$$ 2^n/n! = (2/1)*(2/2)*\ldots*(2/n)$$ $$ \leq 2 * 2/n$$

This quantity can clearly be made arbitrarily small for sufficiently large n, hence the limit is 0 (since $2^n/n!$ is positive)

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We have $$2^n = (1 + 1)^n = \sum_{k = 0}^n\frac{n!}{k!(n-k)!}.$$ Try to conclude from that.

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  • $\begingroup$ Can you give me the ful answer please $\endgroup$ – Badr Eddine Feb 12 at 22:50
  • $\begingroup$ Just look above, there is 2 very good answers. $\endgroup$ – Falcon Feb 12 at 22:51

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