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 $1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.