How to show that lim 2^n/n! = 0 with n--> infinity

  • Thread starter kreil
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  • #1
kreil
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Using the definition of a limit, show that

[tex]\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0[/tex]

If someone could get me started that would be great.

thanks
josh
 
Last edited:

Answers and Replies

  • #2
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[tex]\frac{2^n}{n!}=\frac{2}{n}\frac{2}{n-1}.....\frac{2}{2}
\frac{2}{1}[/tex]

so when [tex]n\rightarrow\infty[/tex]

we get [tex]\frac{2^n}{n!}<\frac{4}{n}[/tex]

to any [tex]\varepsilon > 0[/tex]

[tex]\exists N=[\frac{4}{\varepsilon}]+1[/tex]

when n>N

[tex]\frac{2^n}{n!}<\frac{4}{n}
<\frac{4}{[\frac{4}{\varepsilon}]+1}<\varepsilon[/tex]

now we get the conclusion:

[tex]\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0[/tex]
 
Last edited:

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