Convergence of $1+\frac{1^2\cdot2^2}{1\cdot3\cdot5}+ \frac{1^2\cdot2^2\cdot3^2}{1\cdot3\cdot5\cdot7\cdot9}+...$ I am trying to use the ratio test, for that, I need the general formula for the series.
The general formula for the numerator is $(n!)^2$
The denominator is a sequence of odd numbers that grows by two terms every time but how do I represent it?
Also, any tips for how I can guess the series from a sequence would be greatly appreciated.

Lets try writing the general term
$$a_n=\frac{(n!)^2}{1\cdot 3\cdot 5\cdots (4n-5)(4n-3)}\\a_{n+1}=\frac{((n+1)!)^2}{1\cdot 3\cdot 5\cdots (4n-1)(4n+1)}\\\frac{a_{n+1}}{a_n}=\frac{((n+1)!)^2}{1\cdot 3\cdot 5\cdots(4n-1)(4n+1)}\cdot\frac{1\cdot 3\cdot 5\cdots(4n-5)(4n-3)}{(n!)^2}\\\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(4n-1)(4n+1)}$$
also you can notice that
$$1\cdot 3\cdot 5\cdots (2k-1)(2k+1)=\frac{(2k+1)!}{2^kk!}$$
So $$a_n=\frac{(n!)^2(2k-2)!2^{k-2}}{(4k-3)!}$$
Doing the ratio test should give the same result as above.
