Calculate the limit of the sequence $lim_{nrightarrowinfty} a_n, ngeqslant1$

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) $\frac{1}{2}$

Answer

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