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

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SUMMARY

The limit of the sequence defined by the expression \(\lim_{n \rightarrow \infty} \frac{2^n}{n!} = 0\) is established using the definition of limits. By demonstrating that \(\frac{2^n}{n!} < \frac{4}{n}\) for sufficiently large \(n\), it is shown that this expression can be made less than any arbitrary \(\varepsilon > 0\) by choosing \(N = [\frac{4}{\varepsilon}] + 1\). Thus, the limit converges to zero as \(n\) approaches infinity.

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Using the definition of a limit, show that

\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0

If someone could get me started that would be great.

thanks
josh
 
Last edited:
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\frac{2^n}{n!}=\frac{2}{n}\frac{2}{n-1}...\frac{2}{2}<br /> \frac{2}{1}

so when n\rightarrow\infty

we get \frac{2^n}{n!}&lt;\frac{4}{n}

to any \varepsilon &gt; 0

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

when n>N

\frac{2^n}{n!}&lt;\frac{4}{n}<br /> &lt;\frac{4}{[\frac{4}{\varepsilon}]+1}&lt;\varepsilon

now we get the conclusion:

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

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