Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit? Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit? 
$$ \frac{2^n}{n!} - 0 = \frac{2^n}{n!} < \frac {2^n}{n} <\; ? < \epsilon$$
I dont know  how to simplify $\frac{2^n }{ n}$.
I cannot just do $\frac{2^n}{n}<\epsilon$ right ? 
 A: $\frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot\frac{2}{3} \cdot\frac{2}{4} \cdot\frac{2}{5}  \dots \frac{2}{n}  < \frac{2}{1} \cdot \frac{2}{2} \cdot\frac{2}{3} \cdot\frac{2}{4} \cdot\frac{2}{4}  \dots \frac{2}{4} = \frac{8}{6}\cdot(\frac{1}{2})^{n-3}$
A: For $\; n \geq 3 \;$ we have
$$\frac{2^n}{n!} \;\; = \;\; \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{2}{5} \cdot \frac{2}{6} \cdot \;\; \cdots \;\; \cdot  \frac{2}{n} \;\; \leq \;\; \frac{2}{1} \cdot \frac{2}{2} \cdot \left(\frac{2}{3}\right)^{n-2} \;\; \longrightarrow \;\; 0$$
A: HINT:
$$n!\ge (n/2)^{n/2}$$  
so that
$$\frac{2^n}{n!}\le \left(\frac{8}{n}\right)^{n/2}$$
Alternatively, note that 
$$\frac{2^n}{n!}= \left(\frac{2}{1}\right)\,\overbrace{\left(\frac{2}{2}\right)\left(\frac{2}{3}\right)\left(\frac{2}{4}\right)\cdots \left(\frac{2}{n-1}\right)}^{n-2\,\,\text{terms all less than or equal to }1}\,\frac{2}{n}\le \frac4n$$
A: Consider $\sum \frac{2^n}{n!}$ and do the ratio test:
$$\lim_{n\to\infty} \frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}=\lim_{n\to\infty}\frac{2}{n+1}=0$$
The ratio test implies that the sum is convergent which implies that the limit is $0$
A: the series $\sum_\limits{n=0}^{\infty} \frac {x^n}{n!}$ converges for all x.  In fact it equals $e^x$
That means that the sequence $\{\frac {x^n}{n!}\}$ must converge to 0.
A: $$ \frac21<\frac22<\frac23<\frac13<\frac2{15}<...<\frac1n$$
So the above sequence is bounded by $1/n$; but $\{\frac1n\} \to 0 \text{ as } n\to\infty$.
