Find a stage $n(\epsilon)$ such that $|x_n-l|<\epsilon$

Question

Find a stage $n(\epsilon)$ such that $|x_n-l|<\epsilon, \forall n \le n(\epsilon)$ , where $x_n=\frac{2^n}{n!},l=0$ For $\epsilon=1, 0.1,0.01$ and $0.0001$

What I have DONE

take $\epsilon =1$ now $x_n-l|<\epsilon \Rightarrow |\frac{2^n}{n!}-0|<\epsilon=1\\ |\frac{2^n}{n!}|<1 \Rightarrow 2^n\leq n!$

Since $n!\leq (\frac{n+1}{2})^n$

$\therefore 2^n\le (\frac{n+1}{2})^n \Rightarrow n\geq 3$

$\therefore n(\epsilon)=4$ in the case of $\epsilon =1$

My question is how to find $n(\epsilon) $ for any $\epsilon$

Answer 1

Your $n(\epsilon)=3$ does not work for $\epsilon =1$

Note that $$|\frac {2^3}{3!}-0|=4/3 >1$$ so you need to fix that.

Make sure that your inequalities are in the right order.

Answer 2

While there exists $n_0(\varepsilon)$ such that $|2^n (n!)^{-1}|<\varepsilon$ is true for any $n\geq n_0$ and false for $n<n_0$, you need not find this $n_0$. If you can accept anything that is not nearly close, just note that:

$$ \left| \frac{2^n}{n!}\right|=\left| \frac{2.2.2.\, \cdots .2}{1.2.3.\,\cdots.n}\right| = \left| \frac{2}{1}\right|.\left|\frac{2}{2}\right|.\left|\frac{2}{3}\right|\cdots \left|\frac{2}{n}\right|<2\left| \frac{2}{n}\right|<\varepsilon $$

Thus you can pick $n$ greater than $4/\varepsilon$