Limit 2^n/n using epsilon delta

[Dethrone]

Using epsilon delta, prove
$$\lim_{{n}\to{\infty}}\frac{2^n}{n!}=0$$

Doesn't seem too difficult, but I have forgotten how to do it. Obvious starting point is $\forall \epsilon >0$, $\exists N$ such that whenever $n>N,\left|\frac{2^n}{n!} \right|<\epsilon$.

 

[Euge]

Hi Rido12,

Note that if $n \ge 3$, $$n! = n\cdot (n-1)\cdot (n-2)\cdots \cdot 2\ge n\cdot 2\cdot 2 \cdots \cdot 2 = n\cdot 2^{n-2}.$$ Thus, given $\epsilon > 0$, choosing $N > \max\{3,\frac{4}{\epsilon}\}$ will make $$\left|\frac{2^n}{n!}\right| = \frac{2^n}{n!} \le \frac{4}{n} < \epsilon$$ for all $n \ge N$.

 

[Dethrone]

Hi Euge,

That is actually very smart (Cool). Can you clarify how we got the $3$ in $\max\{3,\frac{4}{\epsilon}\}$? Doesn't $\frac{4}{\epsilon}$ work for all epsilon, or have a missed a case?

 

[Euge]

The max piece was meant to ensure that $ N \ge 3$. So if $ n \ge N$, the inequality $ n! \ge n\cdot 2^{n-2} $ is valid.

 

[Dethrone]

The max piece was meant to ensure that $ N \ge 3$. So if $ n \ge N$, the inequality $ n! \ge n\cdot 2^{n-2} $ is valid.

That's very true. Thanks for the help, again! :D