Determine the largest open set to which $f(z)=\sum_{n=1}^{\infty}(-1)^n(2n+1)z^{n}$ can be analytically continued Let $U=B_1(0)$ and $$f:U \rightarrow \mathbb{C},\qquad f(z)=\sum_{n=1}^{\infty}(-1)^n(2n+1)z^{n}.$$
Determine the largest open set to which $f$ can be analytically continued
Remark: I was given following suggestion: Consider $f(w^2)$.
Do not know how to use the suggestion, I would appreciate any suggestions in this exercise and if is possible, the solution.

The series converges for $\;|z|<1\;$ , for example using the $\;n\,-$ th root test. Now, for any $\;|z|=1\implies z=e^{it}\;,\;\;t\in\Bbb R\;$ , we have that
$$\lim_{n\to\infty}|(-1)^n(2n+1)e^{nit}|=\lim_{n\to\infty}(2n+1)=\infty\neq0$$
and thus the series doesn't converge on the unit circle, so the maximal open set where $\;f\;$ can exist as analytic function is the open unit disk $\;|z|<1\;$
