How to compute the series: $\sum_{n=0}^{\infty} (-1)^{n-1}\binom{1/2}{n}$

Question

I'm wondering about how to show compute this series: $$\sum_{n=0}^{\infty}(-1)^{n-1}\binom{1/2}{n}$$

My approach was to use the general formula of the binomial series, which is: $$(1+z)^r=\sum_{k=0}^{+\infty}z^{k}\binom{r}{k}$$

Yet this can't be used because in this case, we have $|z|=1$. Thus, is there any method that I can use for this?

Answer 1

You 'incororate' it into the binomial coefficient, e.g. for $$ (1+z)^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k}z^k $$ It is $(-1)^{k}\frac{(2k-1)!!}{k!2^k}$