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?

Alex · Answer

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 $(-1)^{k}frac{(2k-1)!!}{k!2^k}$

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

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