# Simplify coefficients for power series and check radius of covergence computate.

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 $$n=4$$.
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 $$1$$.