A limit of a sequence

Question

I'm trying to prove the following limit

$$(\frac{2^n}{n!}) \to 0$$

But it seems difficault to me. How can I prove it?

Thanks.

Likely to be a duplicated. Notice that $2^n/n! = 2/1\cdot 2/2 \cdot 2/3 \cdot 2/4 \dots \leq 2\cdot (2/3)^{n-2}$ — SBF, Nov 28 '14 at 09:42

score 2 · Answer 1 · answered Nov 28 '14 at 09:45

\begin{align} a_n =\frac{2^n}{n!} = \frac{2\cdot \ldots \cdot 2}{1\cdot 2\cdot \ldots \cdot n} \end{align} Assume, that $n\geq 4$, then \begin{align} a_n =\frac{2^n}{n!} = \frac21\frac22\frac23\frac24 \cdot\ldots \cdot \frac2n \end{align} Now, all fractions between $\frac24$ and $\frac2n$ are smaller or equal $\frac24$, so you can bound your sequence \begin{align} a_n &=\frac{2^n}{n!} \leq \frac21\frac22\frac23\frac24 \frac24 \cdot\ldots \cdot \frac24\\ &= \frac21\frac22\frac23 \Bigl(\frac24\Bigr)^{n-3} =\frac86\Bigl(\frac24\Bigr)^{n-3} \end{align} Do you know, what happens to $\Bigl(\frac24\Bigr)^{n-3} $ for $n\rightarrow \infty$?

A limit of a sequence

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