Analytic continuation of $\sum z^{2n}$

Question

If I'm not misusing the root test, the convergence radius of $\sum z^{2n}$ is $\lim \sup \sqrt[2n]{1}=1$ (is this correct?).

Now, is there a closed-form expression of $f(z)=\sum z^{2n}$ so that it be analytically continuated beyond this radius of convergence?

Any hint would be appreciated.

Answer 1

Yes, you use the test correctly.

To see the continuation, write $\sum_{n=0}^{\infty} z^{2n}= \sum_{n=0}^{\infty} (z^2)^n$, and recall $\sum_{n=0}^{\infty} t^n = \frac{1}{1-t}$ for $|t|< 1$. Substitute $z^2$ for $t$.

This assumes the sum starts at $0$ if it starts at $1$ or still elsewhere modify accordingly.