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I want to show that $$ a_n = \frac{3^n}{n!} $$ converges to zero. I tried Stirlings formulae, by it the fraction becomes $$ \frac{3^n}{\sqrt{2\pi n} (n^n/e^n)} $$ which equals $$ \frac{1}{\sqrt{2\pi n}} \left( \frac{3e}{n} \right)^n $$ from this can I conclude that it goes to zero because $\frac{3e}{n}$ and $\frac{1}{\sqrt{2\pi n}}$ approaching zero?

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marked as duplicate by Hans Lundmark, amWhy, kingW3, Davide Giraudo, Lord_Farin Feb 12 '15 at 18:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Hint: Can you rewrite the sequence in such a way that you could use the Squeeze Theorem? – Amzoti Dec 17 '12 at 16:03

yes it's fine, maybe a little bit more care is needed for the $(\frac{3e}{n})^n$ term...

but you could simply say that

$\displaystyle \sum_0^{\infty} \frac{3^n}{n!}=e^3$

in particular it converges and hence the terms must go to zero.

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yes, that good. – Stefan Dec 17 '12 at 16:02

Alternatively to Stirling:

We have \begin{align} a_n =\frac{3^n}{n!} \end{align} Now let $n>3$, then \begin{align} 0\leq a_n &=\frac{3^n}{n!} = \frac{3\cdot3 \cdot 3}{1\cdot 2\cdot 3}\cdot \frac{3^{n-3}}{4\cdot 5 \cdot ...\cdot n}=\frac{3\cdot3 \cdot 3}{1\cdot 2\cdot 3}\cdot \frac{3\cdot 3 \cdot ... \cdot 3}{4\cdot 5 \cdot ...\cdot n}\\ &\leq \frac{9}{2}\cdot \frac{3\cdot 3 \cdot ... \cdot 3}{4\cdot 4 \cdot ...\cdot 4} = \frac{9}{2} \cdot \Bigr(\frac{3}{4}\Bigl)^n\rightarrow0 \text{ as } n\rightarrow \infty \end{align} So we have $a_n\rightarrow 0$ as $n\rightarrow \infty$.

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It's easier to show that, for $n\geq 3$, $n!\geq 3! 4^{n-3}$.

So $|a_n|<\frac{3^n}{6\cdot 4^{n-3}} = \frac{9} 2 \left(\frac{3}{4}\right)^{n-3}$

Show that $\left(\frac{3}{4}\right)^{n-3}\to 0$

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