Sum of the derivative of a geometric series, where $f'(z)= \sum_{n=1}^{\infty}{3^n}z^{2n-1}$

Question

I was doing an assignment where I had to evaluate the derivative of a series. $f(z)=\sum_{n=1}^{\infty}\frac{3^n}{2n}z^{2n}\Rightarrow f'(z)=\sum_{n=1}^{\infty}{3^n}z^{2n-1} $

Then I wrote the series as a the picture below shows. However, I don't see how that is a geometric series. The solution provided states this clearly, however I am not able to do this on my own.

Wouldn't this be different, given that it's $z^{2n-2}$ and not $z^{n-1}$?

Any help would be appreciated.

Answer 1

We have $f'(z)=3z \sum_{n=1}^{\infty}(3z^2)^{n-1}=\frac{3z}{1-3z^2}$ for $|z|<1/ \sqrt{3}.$