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.

Marty and Jonas have done a great job of explaining the main idea of solving this problem, I just want to add to what Jonas mentioned about being able to find that $\dfrac{a_{n+1}}{a_n}$.
When expanding binomials, use the binomial theorem:
$$(a + b)^n = a^n + {n\choose 1} a^{n-1}b + {n\choose 2}a^{n-2}b^2 + \cdots + b^n.$$
Using this we can change  the fraction that you had to:
$$\dfrac{2n^{100}}{n^{100} + 100n^{99} + 4950n^{98} + \cdots}.$$
If you notice on the bottom we have a polynomial of degree $100$, this shares the same degree of the monomial on the top, so we can disregard the other $100$ terms after $n^{100}$. This is why the limit as $n$ goes to infinity of $\dfrac{a_{n+1}}{a_n} = 2$ for our ratio test.   Because the coefficient of the highest powered term on the numerator (2) divided by the coefficient of the highest power term in the denominator (1) is equal to 2.
