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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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Well, you have to say something more than the question itself! Luckily, You've got two good answers. –  Gigili May 25 '12 at 17:46
3  
-1 for taking the copy-and-paste approach this far. –  user31373 May 25 '12 at 17:52
1  
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. –  Grozav Alex Ioan May 25 '12 at 18:01

3 Answers 3

up vote 3 down vote accepted

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? –  André Nicolas May 25 '12 at 17:45

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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Note that the quotient rule test is equivalent to D'Alambert's Criterion, which is equivalent to the $n$-th root test (cauchy's criterion). Basically, if one proves a sequence converges by D'Alambert's or Cauchy's criterion, one proves the series converges and is majored by a geometric series. –  Pedro Tamaroff May 25 '12 at 18:44
1  
What do you mean "equivalent"? Sometimes one helps and the other doesn't. For ex., the series $$\sum_{n=1}^\infty a_n \,\,,\,a_n=\left\{\begin{array}, (1/2)^n &\,\,\text{ if n is odd}\\\, (1/3)^n&\,\,\text{ if n is even}\end{array}\right.$$ is convergent, and the n-th root test so proves it (the lim sup of the n-th root is 1/2), but the D'Alembert's test is inconclusive as the limit doesn't exist... –  DonAntonio May 25 '12 at 22:42
    
IF you're taking the $\limsup$ in the root test, it is fair you do so in the quotetient test. By Stolz Cesaro, one can prove that if the limits exist, then they are equal. That's what I mean. –  Pedro Tamaroff May 25 '12 at 23:21
1  
Apply the the lim sup of the ratio test and you'll get $\,\infty\,$ ! In fact, in the ratio test the positive result is with the lim inf...Never mind, you can check Bonar-Khoury's "Real Infinite Series", or page 3 here math.uga.edu/~pete/243series4.pdf , or page 45 of the great little book "Infinite Series", by Hyslop, from where I get the following example: $\,a_n:=2^{-n-(-1)^n}\,$ , for which the nth root test gives convergence but the ratio test is inconclusive, as the lim sum is greater than 1 and the lim inf is less than 1... –  DonAntonio May 26 '12 at 0:14
    
Fair enough! You convinced me! =D (And, yes, I meant $\liminf$) –  Pedro Tamaroff May 26 '12 at 14:42

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