Ratio test: Finding $\lim \frac{2^n}{n^{100}}$ $$\lim \frac{2^n}{n^{100}}$$ as n goes to infinity of course.
I know that the form os $\frac{a_{n+1}}{a_n}$
$$\frac {\frac{2^{n+1}}{(n+1)^{100}}}{\frac{2^n}{n^{100}}}$$
$$\frac{2n^{100}}{(n+1)^{100}}$$
I am not clever enough to evaluate that limit. To me it looks like it should go to zero using the logic that exponents increase much more quickly than twice. Also I feel like I could reduce it down to $\frac{2}{n}$ but that gives an incorrect answer.

In
Prove that $n^k < 2^n$ for all large enough $n$
I showed that
if $n$ and $k$ are integers
and $k \ge 2$
and
$n \ge k^2+1$,
then
$2^n > n^k$.
Set $k = 101$.
Then, for $n \ge 101^2+1 = 10202$,
$2^n > n^{101}$
or 
$\dfrac{2^n}{n^{100}} > n$
so $\lim_{n \to \infty} \dfrac{2^n}{n^{100}} = \infty$.
This obviously shows that
$\lim_{n \to \infty} \dfrac{2^n}{n^k} = \infty$
for any positive integer $k$.
