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 $(1+x)^{-1/2} (1+x)^{-1/2}$.
 A: Consider the sum
\begin{align}
S_{n} = (-1)^{n} \sum_{m=0}^{n} \binom{-1/2}{m} \binom{-1/2}{n-m}.
\end{align}
Using the results:
\begin{align}
\binom{-1/2}{m} &= \frac{(-1)^{m}(1/2)_{m}}{m!} \\
\binom{-1/2}{n-m} &= \frac{(-1)^{n-m} (1/2)_{n-m}}{(n-m)!} = (-1)^{n+m} \frac{(1/2)_{n} (-n)_{m}}{(1)_{n} (1/2-n)_{m}} \\
\end{align}
the series becomes
\begin{align}
S_{n} &= (-1)^{n} \sum_{m=0}^{n} \binom{-1/2}{m} \binom{-1/2}{n-m} \\
&= \frac{(1/2)_{n}}{(1)_{n}} \ \sum_{m=0}^{n} \frac{(1/2)_{m}(-n)_{m}}{m! (1/2-n)_{m}} \\
&= \frac{(1/2)_{n}}{(1)_{n}} \ {}_{2}F_{1}(1/2, -n; 1/2-n; 1) \\
&= \frac{(1/2)_{n}}{(1)_{n}} \ \frac{\Gamma(1/2-n) \Gamma(0)}{\Gamma(1/2) \Gamma(-n)} = \frac{(1/2)_{n} (1/2)_{-n}}{(1)_{n} (0)_{-n}} = 1.
\end{align}
Hence,
\begin{align}
(-1)^{n} \sum_{m=0}^{n} \binom{-1/2}{m} \binom{-1/2}{n-m} = 1.
\end{align}
A: 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,
$$
\sum_{m=0}^n\binom{-1/2}{n-m}\binom{-1/2}{m}=(-1)^n.
$$
