double factorial series problem In the problem the sum is given as $$\sum\limits_{n=1}^\infty \frac{(2n-1)!!}{(2n)!!}$$
and then when I try to solve it using Gauss's test I get
$$\frac{a_{n}}{a_{n+1}}=\frac{2n+2}{2n+1}$$
but in the solution there is given:
$$\frac{a_{n}}{a_{n+1}}=\frac{2n}{2n-1}$$
my reasoning was:
$$\frac{a_{n}}{a_{n+1}}=\frac{\frac{(2n-1)!!}{(2n)!!}}{\frac{(2n+1)!!}{(2n+2)!!}}=\frac{1\cdot3\cdot...\cdot(2n-1) \times 2\cdot4\cdot...\cdot2n\cdot(2n+2)}{2\cdot4\cdot...\cdot2n \times 1\cdot3\cdot(2n-1)\cdot(2n+1)}=\frac{2n+2}{2n+1}$$
I believe that I made a mistake, but I don't know where?

$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2n)!!^2} = \frac{1}{4^n}\binom{2n}{n} $$
but the RHS behaves like $\frac{1}{\sqrt{\pi n}}$, hence the series is diverging. Also without the exact asymptotics:
$$ \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\prod_{k=1}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right) \geq \frac{1}{4n}.$$
