Determine the largest open set to which $f(z)=\sum_{n=1}^{\infty}(-1)^n(2n+1)z^{n}$ can be analytically continued

Question

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.

Answer 1

To see to what sets $f$ can be analytically continued, it would help if we found a closed form expression for $f$. To do that, we use the hint and consider

$$g(w) := f(w^2) = \sum_{n = 1}^{\infty} (-1)^n (2n+1) w^{2n}.\tag{1}$$

Here, we recognise a derivative, which leads us to look at

$$G(w) = \sum_{n = 1}^{\infty} (-1)^n w^{2n+1} = \frac{-w^3}{1+w^2},\tag{2}$$

from which we find

$$g(w) = G'(w) = \frac{2w^4}{(1+w^2)^2} - \frac{3w^2}{1+w^2}.\tag{3}$$

Now set $z = w^2$ to find

$$f(z) = \frac{2z^2}{(1+z)^2} - \frac{3z}{1+z} = -\frac{z^2+3z}{(1+z)^2}.\tag{4}$$

In this form, it is easy to see that $f$ can be analytically continued to $\mathbb{C}\setminus \{-1\}$, and that that is the largest open set to which $f$ can be analytically continued.

Answer 2

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 $\;|z|<1\;$