Answer
The radius of convergence is $1$.
Work Step by Step
Given $$F(x)=\sum_{n=1}^{\infty} n x^{n}$$ Since $$F'(x)=\sum_{n=1}^{\infty} n^2 x^{n-1} $$ Since $a_n = n^2 x^{n-1}$ and $a_{n+1} = (n+1)^2 x^{n}$, then \begin{align*} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{(n+1)^2 x^{n}}{n^2 x^{n-1}} \right|\\ & =\left|x\right| \lim _{n \rightarrow \infty}\frac{(n+1)^2 }{n^2 }\\ &= \left| x \right| \end{align*}
Then $F'(x)$ converges for $$ \left|x\right|\lt 1 \ \ $$
And the radius of convergence is $1$.
