Prove directly that $a_n = 2^n / n!$ converges $a_n = 2^n/n!$ Hint: look at the ratio of successive terms
I understand that I have to show that for sufficiently large n, $|2^n/n!| < \epsilon$. However, I do not know how to rewrite the inequality in terms of $n$. I do not get how the hint can assist me in directly proving this problem.
Thank you.
 A: Note that $0\leq a_n \leq 2$ for every $n\geq0$.
The hint asks you to look at $b_n=\frac{a_{n+1}}{a_n}$.
I then invite you to answer if $b_n$ is less than, greater than or equal to $1$.
Can you finish from here?
Remember that a bounded and monotonic sequence is convergent (see here).
A: As given in the hint:
$$\frac{a_{n+1}}{a_n} = \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} = \frac{2}{n+1} < \frac{1}{2} \mbox{ for } n > 3$$
It follows with $a_0 = \frac{2^0}{0!}=1$ for $n>5$
$$0 < a_n = \frac{a_n}{a_0}= \frac{a_n}{a_{n-1}}\cdot\frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_2}{a_1}\cdot \frac{a_1}{a_0}< \left( \frac{1}{2}\right)^{n-5}\frac{a_4}{a_0}\stackrel{n\rightarrow \infty}{\longrightarrow}0$$
A: When $b_n=\frac{a_{n+1}}{a_n}=\rho\lt1$ we get convergence by comparison with the geometric series $\sum_{n=0}^{\infty}\rho^n$, if memory serves...
We have that, since $b_n=\frac2{n+1}$.
Edit:  Now I see that you didn't mean the series,  but rather the sequence.   The sledgehammer would be that since the series converges,  the sequence goes to zero. 
To prove it "directly":
Since $a_n\lt a_1\cdot\rho^n$, with $\rho\lt1$, we've got it (let $n\to\infty$).
A: See, for $n\ge 6$ we have, $(n-1)!>2^n$
So, $0<\frac{2^n}{n!}<\frac{1}{n}$ for $n\ge 6$
Now, applying Sandwich theorem, we get 
$\lim_{n\rightarrow\infty}\frac{2^n}{n!}=0$
A: Notice that for $n>3$ we have $$0<a_{n+1}=\dfrac{2^{n+1}}{(n+1)!}=\dfrac{2}{n+1}a_n<\dfrac{1}{2}a_n$$using Squeeze theorem we have $$\lim_{n\to\infty}a_n=0$$
