where the series has the nth term given by $$a_n = \frac {1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}$$
I managed to show that $$a_n = \frac {(2n)!} {4^n (n!)^2}$$
But it doesn't really help.. I'd like to compare it with something since the ratio test doesn't work.
The binomial:
$$\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$$ is the middle, and hence largest, term of the binomial expansion of $\left(1+1 \right)^{2n}=2^{2n}=4^n.$ Since there are $2n+1$ terms, this means that $$\binom{2n}{n}\geq\ \frac{4^n}{2n+1}$$
and hence $$a_n=\frac{1}{4^n}\binom{2n}{n}\geq \frac{1}{2n+1}$$
See more approximations for the central binomial coefficients.
Note that indeed the ratio test is inconclusive, as $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 1\,.$$
If you know Stirling's approximation of the factorial, $$ n! \operatorname*{\sim}_{n\to\infty} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$ then you can show that $a_n \operatorname*{\sim}_{n\to\infty} \frac{1}{\sqrt{\pi n}}$ and the series $\sum_n a_n$ thus diverges by theorems of comparison (for series with positive terms).
If you do not but like probabilities, recall that this is exactly $$a_n = \mathbb{P}\{ X = n \}$$ where $X\sim \mathrm{Bin}(2n,1/2)$ is a Binomial random variable with parameters $2n$ and $1/2$. By standard concentration and anticoncentration results, the Binomial distribution is "roughly uniform" (i.e., its probability mass constant is within constant factors) within $\pm \sqrt{n}$ (that is, more or less the order of a standard deviation) of its expectation, and a constant fraction of its probability mass is on this interval. This implies that $$a_n = \Theta(1/\sqrt{n})$$ leading to the same result.
I consider the second method the most fun, but the first is quite useful (if you do not know Stirling's appproximation but want to, here is a great occasion to do so.) And, of course, there are other ways.
Let use Stirling's approximation
$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$
then
$$\frac {(2n)!} {4^n (n!)^2}\sim\frac {\sqrt{4 \pi n}\left(\frac{2n}{e}\right)^{2n}} {4^{2n} (\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n)^2}$$
