1

I just came across an exercise, however I don't know how to find the limit of $$\lim_{n \to \infty} \frac{2^n}{n!}$$ can any body help? Of course this is not homework, I'm only trying out example myself, from https://people.math.osu.edu/fowler.291/sequences-and-series.pdf page 38.

I know that the limit exists, and it is $0$, checked it in on the wolfram, but I don't know how to solve it.

thanks in advance!

Martin Sleziak
  • 53,687
Trex
  • 63

3 Answers3

5

$$a_n:=\frac{2^n}{n!}\implies \frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}=\frac2{n+1}\xrightarrow[n\to\infty]{}0\implies$$

the infinite series

$$\sum_{n=1}^\infty a_n\;\;\;\text{converges}\;\;\implies\;\;a_n\xrightarrow[n\to\infty]{}0$$

Timbuc
  • 34,191
4

Notice that

$$\frac{2^n}{n!} = {{2 *2*2*2*...*2} \over {1*2*3*4*...*n}}=2*{{2*2*...*2} \over {3*4*...*n}} \leq 2*1*1*...*{{2} \over {n}}={4 \over n}$$

So

$$0 \leq \frac{2^n}{n!} \leq {4 \over n}\xrightarrow[n\to\infty]{}0$$

And hence $\lim_{n \to \infty} \frac{2^n}{n!}=0$

Chris
  • 1,075
0

Generalization for $1<a<k<n$: $$ \lim_{n \to \infty} \frac{a^n}{n!} < a^{k-1} \lim_{n \to \infty} \frac{a^{n-k}}{k(k+1) \ldots n} = k^k \lim_{n \to \infty} \big(\frac{a}{k})^n =0 $$ The last step is since $k(k+1) \ldots n >k^{n-k}$

Alex
  • 19,262
  • 1
    That inequality looks pretty suspicious: you cancel one $;2;$ both in numerator and denominator...and then you keep all the rest of factors of the denominator yet all the other twos in the numerator disappear? – Timbuc May 06 '15 at 09:57
  • If $n>1$ then : $\frac{2...2}{1.2.3...n}\leq \frac{4}{n}$ is easier to show and does the job. – Clément Guérin May 06 '15 at 09:57
  • But for example for $;n=3;$ , the inequality says that $$\frac43=\frac{2^3}{3!}<\frac1{3} ....\text{I doubt it. Am I missing something?}$$ – Timbuc May 06 '15 at 09:59
  • Apologies, solution fixed. – Alex May 06 '15 at 13:41