Proof of convergence of $\sum\limits_{n=1}^\infty \frac{(2n)!!}{(2n+3)!!} $ I am interested in proofs of the convergence of the following series:
$$\sum\limits_{n=1}^\infty \frac{(2n)!!}{(2n+3)!!} $$
I know it converges using Raabe's test. However, I am not allowed to use this test nor the Integral test. I've tried several other tests to no avail.
 A: Using the (convexity) inequalities $1-x\leq e^{-x}$ and $x\geq \log(1+x)$ we have:
$$\frac{(2n)!!}{(2n+3)!!}=\prod_{k=1}^n\left(1-\frac{3}{2k+3}\right)\leq\exp\left(-\sum_{k=1}^{n}\frac{3}{2k+3}\right)\leq\exp\left(-\frac{3}{2}\sum_{k=3}^{n+2}\frac{1}{k}\right)\leq\exp\left(-\frac{3}{2}\sum_{k=3}^{n+2}\log\frac{k+1}{k}\right)=\left(\prod_{k=3}^{n+2}\frac{k+1}{k}\right)^{-3/2}=\left(\frac{n+3}{3}\right)^{-3/2}<6\cdot n^{-3/2}$$
that is tight enough to ensure convergence.
A: I'm not sure which definition of the double factorial for even integers you use, but since the two I know differ by a constant factor, $\sqrt{\frac{2}{\pi}}$, that doesn't matter for the convergence.
If you express the double factorials in terms of ordinary factorials, the term of your series is (up to a constant factor)
$$\frac{2^{n+2}(n+2)!\cdot 2^n n!}{(2n+4)!} = \frac{2^{2n+2}}{\binom{2n+2}{n}}\cdot \frac{1}{(2n+3)(2n+4)}.$$
Since
$$\frac{1}{2^{2n+2}}\binom{2n+2}{n} \sim \frac{1}{\sqrt{\pi n}},$$
the term of your series is $O(n^{-3/2})$, hence the series is convergent.
