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) $\frac{1}{2}$
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) $\frac{1}{2}$
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, $\lim\limits_{n \to \infty} a_n =0$.
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 $\lim\limits_{n\rightarrow\infty}(3/4)^{n-3}=0$.
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 $\,\displaystyle{\frac{3^n}{n!}\to 0}\,$