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?
Convergence of Sequence with factorial
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0Hint: Can you rewrite the sequence in such a way that you could use the Squeeze Theorem? – 2012-12-17
3 Answers
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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0yes, that good. – 2012-12-17
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 $n\rightarrow \infty$.
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 $\left(\frac{3}{4}\right)^{n-3}\to 0$