Find the radius of convergence and coefficients of the power series for function $$F(z) = \sum^\infty_{n=1} \frac{z^n}{(1-z^n)^2} = \sum^\infty_{n=1} f_n(z) $$ for $z\in \mathbb{C}$.
sketch: Power series:
$\sum^\infty_{n=1} \frac{z^n}{(1-z^n)^2} = \sum^\infty_{n=1} \sum^\infty_{i=1} i z^{in} =\sum^\infty_{n=1} a_n z^n$.
I am no sure how to represent simplify these coefficients.
Radius: Function $f_n(z)$ are homomorphic for disk and series $F(z)$ are almost convergence on this disk. So this series represented homomorphic function. Imply that radius of convergence searching power series is $R \leq 1$ When we used simple case where $z=1$ that conditions of convergence does not happen. We have $R(F) = 1$.
Please check my solution and maybe you have idea how to simplify these coefficients?
$a_n = $ sum of divisors of $n$ (including $1$ and $n$).
Think about it. For, e.g., $n=4$ you have the possibilities $(1,3),(2,2),(3,1)$ for $(n,i)$. Summing up the second components is exactly the sum of divisors of $n=4$.
See here: https://en.wikipedia.org/wiki/Divisor_function
And since $1\le a_n\le 1+2+\ldots+n = \frac{n(n+1)}2\le n^2$, and thus $1\le\sqrt[n]{a_n}\le\sqrt[n]{n^2}$, it follows that the radius of convergence is $1$.