What is the radius of the convergence of $\sum_{1}^{\infty} n^{3} \cdot x^{n}$?

Question

I assume that I should use the root test which gives, that $n^2 \cdot |x|$ how can I continue it?

Answer 1

I don't think $n^2$ should be there.

$$\lim_{n \to \infty} |n^\frac{3}{n} x|=|x|<1$$

Answer 2

With root test, you will have $n^{3/n}|x|\rightarrow|x|$. We now let $|x|<1$.

Answer 3

By the action of the forward difference operator, $$ (1-x)^4 \sum_{n\geq 1} n^3 x^n $$ is a polynomial, hence the radius of convergence is one.