Proving Convergence without Ratio Test Without using the Ratio test, how would one prove the convergence (which is intuitive) of this summation?
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n!}{1\cdot3\cdot5\cdots(2n-1)}$$
My Approach:
I suspect that I must convert the series in the bottom as a fraction of factorials to eliminate the evens.
To do this, I will convert $1\cdot3\cdot5\cdots(2n-1)$ into $\frac{(2n)!}{2^n(n!)}$ which divides the evens from $(2n)!$.
This leaves the sum as
$$\sum_{n=1}^\infty\frac{2^n(n!)^2}{(2n)!}$$
We can set up an inequality such that $a_{n}\geq{a_{n+1}}$:
Note that this is more of a hunch on intuition. Is there a better way for me to derive this assertion?
$$\frac{2^n(n!)^2}{(2n)!}\geq\frac{2^{n+1}((n+1)!)^2}{(2n+2)!}$$
$$\frac{2^n(n!)^2}{(2n)!}\geq\frac{2(2^n)(n+1)^2(n!)^2}{(2n+2)(2n+1)(2n)!}$$
$$\frac{2(n+1)^2}{(2n+2)(2n+1)}\geq1$$
$$\frac{n^2+2n+1}{2n^2+3n+1}\geq1$$
which holds for some value of $n\gt1$
The limit of the original series is where I have trouble with (in retrospect, I should have done this first to see if the limit even converges!) 
So how would one go about determining convergence without Ratio test and certainly without higher level mathematics? Is my sequence of steps on point? How could I make this "more formal"? And how can I take the limit of the the equation inside the sum?
 A: Using Euler's Beta function,
$$S=\sum_{n\geq 1}\frac{(-1)^{n-1}2^n n!^2}{(2n)!}=-\sum_{n\geq 1}n(-2)^n B(n,n+1)=-\int_{0}^{1}\sum_{n\geq 1}n(-2)^n x^{n-1}(1-x)^n\,dx $$
leads to:
$$ S = 2\int_{0}^{1}\frac{1-x}{(2x^2-2x-1)^2}\,dx=\int_{0}^{1}\frac{dx}{(2x^2-2x-1)^2} $$
and by partial fraction decomposition, or the substitution $x=t+\frac{1}{2}$ followed by integration by parts, we get:
$$\boxed{\, S = \sum_{n\geq 1}\frac{(-1)^{n-1} n!}{(2n-1)!!}=\color{red}{\frac{1}{3}+\frac{1}{3\sqrt{3}}\,\log\left(2+\sqrt{3}\right)}\approx 0.586782.\; }$$
Hence yes, it is convergent, and convergent to $\frac{1}{3}+\frac{\log(2+\sqrt{3})}{3\sqrt{3}}$.
A: For $n\geq 2$, 
$$
 \frac{ n!}{1\cdot3\cdot5\cdots(2n-1)}\leq \frac{n!}{1\cdot 2 \cdot 4\cdots (2n-2)}=\frac{n!}{2^{n-1}(n-1)!}=\frac n{2^{n-1}}\leq 10\cdot\frac{1.1^n}{2^{n-1}}.
$$
A: Probably less easy than Jack D'Aurizio's answer, we could identify that 
$$\sum_{n=1}^\infty(-1)^{n-1}\frac{(n!)^2}{(2n)!}x^n$$ is the Taylor expansion of $$\frac{x}{x+4}+\frac{4 \sqrt{x} }{(x+4)^{3/2}}\,\sinh
   ^{-1}\left(\frac{\sqrt{x}}{2}\right)$$ from which follows the result.
