Determing Mclaurin series for $\sqrt{1-z^2}$.

Question

Determine the Mclaurin series and the convergence radius for $\sqrt{1-z^2}$.

I tried to adapt the series $(1+z)^\alpha=\sum_\limits{n=1}^{\infty} {{\alpha}\choose {n}}z^n$.

$\sqrt{1-z^2}=1+\sum_\limits{n=0}^{\infty} {{\frac{1}{2}}\choose {n}}z^n=1+\sum_\limits{n=0}^{\infty} (-1)^n\frac{(2n-3)!}{2n!!}z^{2n}$.

The solution on the book is $1-\frac{z^2}{2}-\sum_\limits{n=2}^{\infty} \frac{(2n-3)!}{2n!!}z^{2n}$

Question:

What did I do wrong? Can someone help me solve the exercise?

Thanks in advance!