complex analysis - Uniform convergence of $sum_{n=0} ^infty frac{(-1)^n}{2n+1} left(frac{2z}{1-z^2}right)^{2n+1}$

I am given:
$$f(z)=\sum_{n=0} ^\infty \frac{(-1)^n}{2n+1} \left(\frac{2z}{1-z^2}\right)^{2n+1}$$
And asked to show that this complex function series converges uniformly for $|z|\leq \frac{1}{3}$. I am also asked to determine its complex derivative in simplified form.

$1)$ I try:

$$f'(z)=\frac{2z^2 +2}{(1-z^2)^2} \cdot \sum_{n=1} ^\infty \frac{(-1)^n}{2n+1} \cdot (2n+1) \left(\frac{2z}{1-z^2}\right)^{2n}$$
Here I pulled out the factor from the chain rule (+quotient rule) as it does not depend on $n$.
$$f'(z)= \frac{2z^2 +2}{(1-z^2)^2} \cdot \sum_{n=1} ^\infty (-1)^n \left(\frac{2z}{1-z^2}\right)^{2n}$$
However, I seem to now be missing the factor of $2n$ in the enumerator as I wanted it to be a complex cosine, what went wrong here? I also do not have the right index yet, I suspect this may be related to my problem.

$2)$ For convergence I recognise this as a power series so I can use one of the many tests, I will use the ratio test:

$$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty} \left|-\frac{2n+1}{2n+3} \left(\frac{2z}{1-z^2}\right)^2\right|=\frac{4 |z|^2}{|1-z^2|^2}$$
For convergence we need $\frac{4 |z|^2}{|1-z^2|^2} <1$

Observe by reverse triangle inequality $$\frac{2 |z|}{|z^2-1|} \leq \frac{2 |z|}{||z|^2-1|}$$
Now since $|z|\leq 1/3$, we have $1\geq ||z|^2 -1| \geq \frac{8}{9}$ or $1\leq \frac{1}{||z|^2 -1|} \leq \frac{9}{8}$ therefore
$$\frac{2 |z|}{|z^2-1|} \leq 2 \cdot 1/3 \cdot 9/8 =3/4$$
The square of this is expression is what we are interested in, this is $9/16<1$

Answer

Observe for $|z|\leq \frac{1}{3}$ We have:

$$|f_n|=\left| \frac{(-1)^n}{2n+1} \left(\frac{2z}{1-z^2}\right)^{2n+1} \right| = \frac{1}{2n+1} \left|\left(\frac{2z}{1-z^2}\right)\right|^{2n+1}$$
By reverse triangle inequality we get:
$$|f_n|\leq \left|\left(\frac{2|z|}{|1|-|z^2|}\right)^{2n+1} \right|=\left|\left(\frac{2|z|}{|1|-|z|^2}\right)^{2n+1} \right| \leq \left(\frac{2 / 3}{1-1/9}\right)^{2n+1}=(\frac 3 4)^{2n+1} = \frac{3}{4} \cdot \left(\frac{9}{16}\right)^n= M_n$$
By the Weiserstrass M-test $\sum M_n$ is a convergent infinite geometric series hence the original sum is convergent in the ball of radius $\frac 1 3$.