I would like to compute the sum $\displaystyle\sum_{n=0}^\infty\dfrac{(2n-3)!!}{(2n)!!},$ where the double exclamation point refers to double factorial.
Using double factorial identities we get the ratio of the $(n+1)$th term to the $n$th as $\dfrac{2n-1}{2n+2}\to 1,$ so the ratio test is inconclusive. Also it is not alternating.
We can write the $n$th term as $$\dfrac{1}{2n}\cdot\dfrac{2n-3}{2n-2}\cdot\dfrac{2n-5}{2n-4}\cdot\dotsb\cdot\dfrac{3}{4}\cdot\dfrac{1}{2}$$ so we see that it looks like a harmonic series multiplied termwise by an increasing number of fractions between 1/2 and 1.
Wolfram alpha just reports that ratio test fails, but Mathematica claims it converges and its value is $0$, which is expected since this is the power series for $\sqrt{1-z}$ evaluated at $z=1$. How can we show convergence?
$$a_n=\frac{(2n-3)!!}{(2n)!!} = \frac{1}{(2n-1)4^n}\binom{2n}{n} = O\left(\frac{1}{n\sqrt{n}}\right), $$ hence the series is convergent.
To find its value, we may recall that by the extended binomial theorem: $$ \sum_{n\geq 0}\frac{x^n}{4^n}\binom{2n}{n} = \frac{1}{\sqrt{1-x}}, $$ hence it follows that: $$ \sum_{n\geq 0}\frac{x^{2n}}{4^n(2n-1)}\binom{2n}{n} = -\sqrt{1-x^2} $$ and our sum is just that series for $x=1$, hence zero.
That also follows by noticing that $a_n$ can be written in a telescopic fashion: $$ \frac{(2n-1)!!}{(2n)!!}-\frac{(2n-3)!!}{(2n-2)!!} = \frac{(2n-1)!!-2n\cdot(2n-3)!!}{(2n)!!} = -a_n.$$