Determine the radius of convergence and find the coefficients of the power series which is equivalent to $$F(z) = \sum^\infty_{n=1} \frac{z^n}{(1-z^n)^2}$$ for $z\in \mathbb{C}$.
That is we would like to find such complex coefficients $a_1,a_2,...$ that $F(z)=\sum^\infty_{n=1} a_n z^n$. Having those, we would find the radius from the Cauchy-Hadamard formula.
The function $F$ is definitely analytic on $\mathbb C\setminus\mathbb T$, where $\mathbb T$ denotes the unit circle. Since $\lim_{z\to 1}F(z)$ does not exist, $F$ cannot be extended. So, $F(z) = \sum_{k=0}^\infty a_kz^k$ for $|z|<1$. To obtain the $a_k$'s, note that $\frac 1{1-z^n} = \sum_{k=0}^\infty z^{kn}$.