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.

I come here with my fancy way: look at the series
$$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{n^{100}}{2^n}$$
Since this is a positive series we can apply the quotient rule, say:
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{100}}{2^{n+1}}\frac{2^n}{n^{100}}=\frac12\left(\frac{n+1}n\right)^{100}\xrightarrow[n\to\infty]{}\frac12<1$$
The above means this positive series converges, and this means 
$$\;\displaystyle{\frac{n^{100}}{2^n}\xrightarrow[n\to\infty]{}0\implies\frac{2^n}{n^{100}}\xrightarrow[n\to\infty]{}\infty}\;$$
