Is sequence limited and what is limit

Question

I am stuck at one problem. So I have to check if sequence is convergent.

$$\frac{2^x}{x!}$$

My thinking was to calculate limit and if limit exists it's convergent, but I am struggling with this:

$$\lim_{x \to \infty} \frac{2^x}{x!} = ?$$

I can't do it with L'Hopital's rule or with logarithm. Does that mean that this sequence is divergent?

Answer 1

You could note that for $n\ge 4$ you have $$\frac{2^n}{n!} = \frac2n \frac{2^{n-1}}{(n-1)!} \le \frac12 \frac{2^{n-1}}{(n-1)!}.$$ Thus by induction you get that for the $n$-th term $$\frac{2^n}{n!}\le \frac{1}{2^{n-3}} \frac{2^3}{3!}=2^{-n} \ 64/6.$$ And, I assume you can finish this.

Answer 2

I take it from your notation that x is a positive integer. let $U_x$ be your sequence.

Then suppose x>3 :

x!=1*2*3*..*x > $2*3^{x-2}$

-> 0 < $U_x < \frac{2^x}{2*3^{x-2}} = 3*(\frac{2}{3})^{x-1}$

=> ($U_x$) ->0 , when x->∞

Answer 3

We know that$$e^x=\sum\dfrac{x^n}{n!}.$$ So we have $$e^2=\sum\frac{2^n}{n!}\implies \lim_{n\to\infty}\frac{2^n}{n!}=0$$