would the sum from 1 to infinity of $\frac{n^3}{(\ln{2})^n}$ converge? In the limit n tends to infinity the denominator grows more quickly and so the terms go to zero.
Using the ratio test I get $\frac{(n+1)^3}{n^3\ln{2}}$ but im not sure if $(n+1)$ grows fast enough to make the ratio greater than one.
On
No, because ln(2) is less than one, so the terms clearly are all positive and tend to infinity, so the sequence diverges.
But for any value of x bigger than e, this is true, as we then have $\sum_{n=1}^{\infty} \frac{x^3}{a^x}$ (where $a = ln(x)>1$), and we can use the ratio test to get $\left(\frac{x+1}{x}\right)^3/a$. This limit tends to 1/a by L'hopital's rule, so we know by the ratio test that this sequence converges.
Observe that $\ln 2\simeq 0.69$, and in particular $\ln 2 \in (0,1)$. Therefore, $\frac{n^3}{(\ln 2)^n} > n^3$ (actually much bigger), and you can conclude easily that the series diverges.