Finding the lim sup and lim inf of a sequence?

Question

Let $a_n=1$ if $n=2^k$ for some positive integer $k$, and $a_n=\frac{1}{n!}$ otherwise.

Find $\limsup_n a_n$ and $\liminf_n a_n$.

Find $\limsup_n \frac{|a_{n+1}|}{|a_n|}$.

Find $\limsup_n |a_n|^{1/n}$.

Attempt at a solution: For the first part, since all $n,k>0$, $a_n<0$ whenever $n\not=2^k$, so the $\limsup = 1$. For the lim inf, would if be 0, since the $\frac{1}{n!}$ portion decreases as $n\to \infty$?

For the second part, I was thinking that this might occur when $a_{n+1}=1$, which would make $a_n$=$1/(2^k -1)!$, so would this be $(2^k-1)!$ ?

For the last portion, $\limsup_n |a_n|^{1/n}$, would this occur when $a_n=1$ so $1^{1/2^k}=1$?

Thank you in advance!

Answer 1

Recall that $\lim \sup_n a_n$ is an abbreviation for $\lim_{n\to \infty} \sup \{a_j :j>n\}$. So in each case consider $\sup \{a_j :j>n\}$ for any $n$ and it should be easy.

Answer 2

It seems to me that in the second part, the lim sup is infinity, since as you mention we have the value $(2k-1)!$, but $k$ can be arbitrarily high.