The question is:
let $f(z)=\sqrt{z}$, which is considered analytic in $z \in \mathbb{C}\setminus [0,\infty).$ Find the taylor expansion of $\sqrt{z}$ at $z=-1$ and determine the radius of convergence.
Now i understand that the radius of convergence can easily be considered to be the point from the origin to the nearest singularity or by using the limsup but either way this all requires me figuring out the power series of $\sqrt{z}$, can someone please tell me if i'm correct so far
$$\sqrt{z} = (z)^{\frac{1}{2}}=((1+z)-1)^\frac{1}{2}$$ which we can recognise as the bionomial theorem with fractional power. then the power series is $$\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}(z+1)^{n} = $$ $$\sum_{n=0}^{\infty}\frac{(-1)^{n-1}(2n-3)!!}{n!2^n}(z+1)^{n} =$$ $$\sum_{n=0}^{\infty}\frac{(-1)^{n-1}2^{n-1}(n-1)!(2n-3)!!}{n!2^{2n-1}(n-1)!}(z+1)^{n} =$$ $$\sum_{n=0}^{\infty}\frac{(-1)^{n-1}(2n-2)!!(2n-3)!!}{n!2^{2n-1}(n-1)!}(z+1)^{n} =$$ $$\sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{n2^{2n-1}}\frac{(2n-2)!}{(n-1)!^{2}}(z+1)^{n} =$$ $$\sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{n2^{2n-1}}\binom{2n-2}{n-1}(z+1)^{n} =$$
with $a_n = \frac{(-1)^{n-1}}{n2^{2n-1}}\binom{2n-2}{n-1}$
then it converges for $|z+1| < 1$
the problem is my intuition tells me i've made a mistake and i dont know where. any help would be appreciated.