Nature of Singularity for $f(z)=\sum_0^\infty \left(\frac{1-z}{1+z}\right)^n$

Question

Consider the function $$f(z)=\sum_0^\infty \left(\frac{1-z}{1+z}\right)^n$$ It has an isolated singular point at $z=-1$. I'm trying to find the nature of this singular point. To do this, I have to do Laurent expansion of this around $-1$. I don't know how to do this. But how about an expansion in region $|z|<1$. Using the binomial expansion $$\frac{1}{1-\zeta}=\sum_{k=0}^\infty \zeta^k$$ is valid provided that $|\zeta|<1$. We can write $$\frac{1}{1+z}=\sum_{k=0}^\infty (-1)^k z^k$$ So that
$$f(z)=\sum_0^\infty (1-z)^n\left(\sum_{k=0}^\infty (-1)^kz^k\right)^n=\sum \left[(1-z)-(1-z)z+(1-z)z^2-\cdots \right]^n$$ $$f(z)=\sum(1-2z+2z^2-2z^3+\cdots)^n$$ We can't use it to find the nature of singularity. Please help me with this.

Answer 1

$f$ is not even defined in any deleted neighborhood of $-1$ because $|\frac {1-z} {1+z} | >3$ for $0 <|1+z| <1/2$. The series defining $f$ is divergent.

The series defining $f$ converges when $Re (z) >0$. Also, it converges uniformly on compact subsets of the right half plane. Hence, it has no singularites there.