Sum of products of binomial coefficient $-1/2 \choose x$ I am having trouble with showing that
$$\sum_{m=0}^n (-1)^n {-1/2 \choose m} {-1/2 \choose n-m}=1$$
I know that this relation can be shown by comparing the coefficients of $x^2$ in the power series for $(1-x)^{-1}$ and $(1+x)^{-1/2} (1+x)^{-1/2}$.

Using power series,
$$
(1+x)^{-1/2}=\sum_{m=0}^\infty \binom{-1/2}{m}x^m
$$
and then power series multiplication, one obtains
$$
(1+x)^{-1}=\sum_{k=0}^\infty \binom{-1/2}{k}x^k\sum_{m=0}^\infty \binom{-1/2}{m}x^m
=\sum_{n=0}^\infty x^n\sum_{m=0}^n\binom{-1/2}{n-m}\binom{-1/2}{m}
$$
so that, comparing coefficients of equal degree,
$$
\sum_{m=0}^n\binom{-1/2}{n-m}\binom{-1/2}{m}=(-1)^n.
$$
