$\lim_{n\to \infty} n^3(\frac{3}{4})^n$

If $\frac {a_{n+1}} {a_n} \to l$ with $|l|<1$ then $a_n \to 0$. (In fact the series $\sum a_n$ converges).

In this case $\frac {a_{n+1}} {a_n} =(1+\frac 1 n)^{3}(\frac 3 4) \to \frac 3 4$.

There are a number of ways to approach this problem. As others have mentioned, the convergence of $\displaystyle \sum^\infty n^3(3/4)^n$ implies $n^3(3/4)^n \to 0$ as $n \to \infty$, and it looks like this is what you're trying to exploit.

Exercise: If $\displaystyle \lim_{n \to \infty} a_n \neq 0$, prove that $\displaystyle \sum_{n=1}^\infty a_n$ cannot converge.

I'm a fan of logarithms myself, so here is another approach:

We have $\ln \Big( n^3(3/4)^n \Big) = 3\ln(n) + n \ln(3/4)$. Notice that $3/4 < 1$, so its natural log will be negative; we can rearrange the expression to emphasize this:

$$3\ln(n) + n \ln(3/4) \ = \ 3 \ln(n) - n \ln(4/3)$$

At this point, we note that the linear term grows faster than the logarithmic term, so we see that $\ln(a_n) \to - \infty$, and because the natural logarithm is a continuous function, we can conclude that $a_n \to 0$.

If you feel that the "grows faster" argument is hand-wavy, you can make it rigorous by taking a derivative, wherein you'll find that it's negative for all $x$ beyond $x \approx 10.4$ and tending to $-\ln(4/3)$, implying unbounded growth toward $- \infty$.