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Calculate the limit of the sequence

$\lim_{n\rightarrow\infty}\ a_n, n\geqslant1 $

knowing that

$\ a_n = \frac{3^n}{n!},n\geqslant1$

Choose the right answer:

a) $1$

b) $0$

c) $3$

d) $\frac{1}{3}$

e) $2$

f) $\frac{1}{2}$

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    I'm sorry for the copy and paste approach.. I keep trying but my results seem to go nowhere near the answers. I must finish the exercise book before my university admission exam and this is the only website where people can help me. I really appreciate how amazing the community is.2012-05-25

3 Answers 3

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Using D'Alambert's criterion, we can see that

$ \lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty} \frac{3^{n+1}n!}{3^{n}(n+1)!}= \lim_{n \to \infty} \frac{3}{n+1}=0$

Thus, $\lim\limits_{n \to \infty} a_n =0$.

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Hint: $0\le{3^n\over n!}= {3\over 1}\cdot{3\over 2}\cdot {3\over3}\cdot \underbrace{{3\over4}\cdot{3\over5}\cdots\cdot{3\over n-1}\cdot {3\over n}}_{\le (3/4)^{n-3}}, $ and $\lim\limits_{n\rightarrow\infty}(3/4)^{n-3}=0$.

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    Upvote of course. But as someone who has never given a multiple choice question, and never will, I wonder about strategy. Is it best to respond mechanically by going to the nearest book item that may fit, or should one actually think?2012-05-25
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This is a "fancy" way to find the limit:

1) Show that $\,\displaystyle{\sum_{n=1}^\infty \frac{3^n}{n!}}\,$ converges (for example, by the quotient rule test, or the n-th root test)

2) Deduce $\,\displaystyle{\frac{3^n}{n!}\to 0}\,$

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    Fair enough! You convinced me! =D (And, yes, I meant $\liminf$)2012-05-26