Sum of products of binomial coefficient $-1/2 \choose x$

Question

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 $(1+x)^{-1/2} (1+x)^{-1/2}$.

Answer 1

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, $$ \sum_{m=0}^n\binom{-1/2}{n-m}\binom{-1/2}{m}=(-1)^n. $$

Answer 2

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}