Let $p>0$. Show the $\lim_{n\to\infty}(a^n/n^p)$ is equal to: $\lim_{n\to\infty}   (a^n/n^p) = 
\begin{cases}
0             &\text{if }|a|\le1,\\
\infty      &\text{if }a > 1,\\
\text{D.N.E}         &\text{if }a < -1.
\end{cases}$
I've proved |a| part and I'm going to be proving $a < -1$ shortly. The problem I'm having is with $a > 1$. A hint for the problem says to use a lemma from a previous question: (If $L>1$, then $\lim_{n\to\infty}(|s_n|)=\infty$). I just really don't know how to use it properly. I got the limit into indiscriminate form, set my function equal to $|s_n|$ and stated the lemma, concluding the "proof". I'm fairly certain I've proven nothing. Hoping someone could help me out here... -_-
 A: Do not worry about the downvote. The answer is correct
You can use the result

if $\lim_{n\to \infty } \frac{a_{n+1}}{a_n} = b $ and $b> 1 $, then $a_n \longrightarrow_{n\to \infty} \infty$.

A: I can attempt to prove this case a > 1. Since lim ((n+1)/n)^2p = 1 as n --> infinity for a given p > 0. There exists an N such that n > N ==> ((n+1)/n)^2p < a. Thus we can prove by induction on n > N that a^n > n^(2p). Thus a^n/n^p > n^p --> infinity when n --> infinity and we're done.
A: My other answer is: Use the ratio test (a^(n+1)/(n+1)^p / a^n/n^p) = a(n/n+1)^p --> a as n--> infinity. So comparing this series with the geometric series whose r = a > 1. This series goes to infinity. I think we have the anser.
A: This is the 3rd proof. Let m =[p] + 1 then a^x/x^p > a^x/x^m for all x > 1. Using L'hopital rule we have:
lim x --> infinity (a^x/x^m) = lim x --> infinity (a^x*lna/mx^(m-1)) = ...= lim x--> infinity (a^x(lna)^m/m!) = infinity. So lim n --> infinity (a^x/x^p) = infinity,and lim n--> infinity (a^n/n^p) = infinity !!
