Show that $\lim(\frac{2^n}{n!}) = 0$
The only tool I can use is the definition of a limit. Here is what I have so far, but I am out of ideas as to how to get rid of the $2^n$.
$$|\frac{2^n}{n!} - 0| = 2^n \frac{1}{n!} < 2^n \frac{1}{2^n} < 2^n \frac{1}{n}$$
I get the right part of the inequality by observing that when $n > 3$, then $n! < 2^n$ so $\frac{1}{n!} < \frac{1}{2^n} < \frac{1}{n}$.
Let $\epsilon > 0$ and choose $k \in \mathbb{N}$ such that $k > \epsilon$...
Am I even going in the right direction here by trying to get a $\frac{1}{n}$ by itself? Is there a clever way to pick an $\epsilon$ here?
edit: Here is my work after listening to the comments
Basically I just need to conclude that $\lim(\frac{2}{3})^n = 0$ and then the rest of the problem follows.
Let $\epsilon > 0$ and $k \in \mathbb{N}$ such that $k > \epsilon$. Let $n \geq k ~~\forall n \in \mathbb{N}$. Then when $n \geq 1$ and $k = 1$, the inequality is true. Thus $$ \frac{2}{3}^n < \epsilon < 1$$ and $\lim(\frac{2}{3})^n = 0$.
$\endgroup$ 43 Answers
$\begingroup$A bit late answer but I think worth mentioning it.
Your idea to split off $\frac 1n$ works quite well. For $n>3$ you have
$$\frac{2^n}{n!}= \frac{2\cdots 2}{1\cdots n} = 4\frac{\overbrace{2\cdots 2}^{(n-3)\; factors}}{\underbrace{3 \cdots (n-1)}_{(n-3)\; factors} \cdot n}< \frac 4n$$
Now you can choose $N_{\epsilon}$ easily for any given $\epsilon >0$.
$\endgroup$ $\begingroup$Here's another, albeit indirect, way to show that $\displaystyle \lim_{n\to\infty}\frac{2^n}{n!}=0$.
Consider the infinite series $\displaystyle \sum_{n=0}^{\infty}\frac{2^n}{n!}$. By the Ratio Test we can easily show that this series converges. Since the series converges, its terms approach zero.
If it feels like cheating… well, that's because all the hard work of comparing with an appropriate geometric sequence/series was done in the proof of the Ratio Test.
$\endgroup$ $\begingroup$Let $x_n = \frac{2^n}{n!}$. Observe that $x_7 = \frac8{315} < \frac{128}{2187} = \left(\frac23\right)^n$, and that if $x_n < \left(\frac23\right)^n$ for some integer $n\geqslant 7$ then $$ x_{n+1} = \frac{2^{n+1}}{(n+1)!} = \frac{2^n}{n!}\cdot \frac2{n+1} = x^n\cdot \frac2{n+1} < \left(\frac23\right)^n\cdot\frac23 = \left(\frac23\right)^{n+1}, $$so that $x_n<\left(\frac23\right)^n$ for all $n\geqslant 7$. Let $\varepsilon>0$ and choose a positive integer $N>\frac{\log\varepsilon}{\log2/3}$. Then $$x_n<\left(\frac23\right)^n <\varepsilon,\quad n\geqslant N, $$and hence $\lim_{n\to\infty} x_n=0$.
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