Obtain the region of convergence and the sum of the series $f(z) = \sum _{n=1}^{\infty }\:\frac{1}{n}\left(1-z^2\right)^n$. Study the uniform convergence over compacts and represent $f(z)$ as a power series around $z=1$
What I did was substitute $(1-z^2)^n = x$ and work with $\sum _{n=1}^{\infty }\:\frac{1}{n}\left(x\right)^n$ as a power series, which I found that it converged to $- \ln(z^2)$ for $x \in [-1,1)$.
I don't really know how to conclude anything about the uniform convergence over compacts, and as for the power series representation around $z=1$ I just know that I need to write $\sum _{n=1}^{\infty }\:\frac{1}{n}\left(1-z^2\right)^n$ as $\sum _{n=1}^{\infty }\:\ c_n\left(z-z_0\right)^n$ for $z_0 = 1$ and some sequence $(c_n)$ in $\mathbb{C}$ that I don't know how to find either.
Partial answer: $\sum \frac 1 n (1-z^{2})^{n} =\sum \frac 1 n (2-(1-z))^{n}(1-z)^{n}=\sum \frac 1 n \sum\limits_{j=0}^{n} \binom {n} {j} (-2)^{j} (1-z)^{n-j} (1-z)^{n}$. The coefficient of $(1-z)^{m}$ in this is $\sum\limits_{n=m/2}^{m} \frac 1 n (-2)^{2n-m} \binom {n} {2n-m}$. [When $m$ is odd the sum starts with $n = \frac {m+1} 2$ and whe $m$ is even the sum starts with $n = \frac m 2$].
The series converges uniformly on compact subsets of $\{z: |1-z^{2}| <1\}$.