# Finding sum of the series $\sum_{i=1}^\infty \frac{(-1)^n (2n-1)!!}{(2n)!!}$ [closed]

I would like some help in calculating the following sum, I don't know how to handle double factorial.

The $n$-th term is $$(-1)^n\frac{1\times 3\times5\times\cdots\times(2n-1)} {2\times 4\times6\times\cdots\times(2n)} =\frac{1}{n!}\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)\cdots\left(-\frac{2n-1}2\right).$$ According to the binomial theorem, that is the coefficient of $x^n$ in the power series for $(1+x)^{-1/2}$, etc.