Finding the limit of (3^n)/(n^3) and how to tell if it is convergent or divergent In Calculus 2 we just started on doing sequences and I understand that to find the limit you can use l'hopital's rule and the sandwich theorem and a few other tricks but I'm generally confused on when to use what. 
So we have this problem:   3^n/n^3
and we need to state if it diverges or converges, and if convergent, find the limit. 
I really have no idea how to go about solving this, if someone can point me in the right direction it would be much appreciated. Thanks. 
 A: $$3^n=(2+1)^n=2^n+2^{n-1}\cdot \binom{n}{1}+\cdots+2\binom{n}{n-1}+\binom{n}{n}$$
When $n\ge 4$, there's a term $\binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{4!}$, which is a polynomial of degree $4$ with a positive $n^4$ coefficient. All the other terms of this expansion are positive. Ratio of higher degree to lower degree polynomial with positive highest degree coefficients as $x\to\infty$ is $\infty$.
(you can see this by dividing both the numerator and denominator by $x^k$ ($k$ -- numerator's highest degree))
A: You may observe that
$$
f(x)=x \ln (3/2)-3 \ln x >0, \quad x>24, \tag1
$$ since
$$
f'(x)=\ln (3/2)-\frac3x >0, \quad x>24.
$$ Then you deduce from $(1)$ that
$$
\frac{3^n}{n^3} > 2^n,  \quad n>24,
$$ and, as $n \to \infty$, $\dfrac{3^n}{n^3} \to+\infty$.
A: $$\frac{3^n}{n^3}= \left( \frac{\sqrt[6]{3}^n}{\sqrt{n}} \right)^6$$
Now, by Bernoulli inequality
$$\sqrt{3}^n \geq 1+(\sqrt[6]{3}-1)n>(\sqrt[6]{3}-1)n $$
Therefore
$$\frac{3^n}{n^3}\geq \left((\sqrt[6]{3}-1)\sqrt{n} \right)^6=\left(\sqrt[6]{3}-1\right)^6 n^3$$
