$1+z(z+1)(1+z)^2$ and $\frac{1}{1+z^2}$ as power series

Question

How can one write the functions

$$1+z(z+1)(1+z)^2$$

and

$$\frac{1}{1+z^2}$$

as power series at point $0$ and find out their convergence radius?

I know that

$$\frac{1}{1+z^2} = \sum_{k=0}^\infty (-1)^k \cdot z^2$$

And I've seen that

but that doesn't help me..

Answer 1

In the case of $\displaystyle\sum_{k=0}^\infty(-1)^kz^{2\color{red}{k}}$, you can apply the ratio test. It follows from it that the radius of converrgence is $1$.

On the other hand\begin{align}1+z(z+1)(1+z)^2&=1+z(z+1)^3\\&=1+z+3z^2+3z^3+z^4+0\times z^5+0\times z^6+\cdots,\end{align}whose radius of convergence is $\infty$.