Testing for convergence in Infinite series with factorial in numerator I have the following infinite series that I need to test for convergence/divergence:
$$\sum_{n=1}^{\infty} \frac{n!}{1 \times 3 \times 5 \times \cdots \times (2n-1)}$$
I can see that the denominator will eventually blow up and surpass the numerator, and so it would seem that the series would converge, but I am not sure how to test this algebraically given the factorial in the numerator and the sequence in denominator. The recursive function for factorial $n! = n \times (n-1)!$ doesn't seem to simplify things in this case, as I cannot eliminate the $(2n-1)$ in the denominator. Is there a way to find a general equation for the denominator such that I could perform convergence tests (e.g. by taking the integral, limit comparison, etc.)
 A: Hint: Use the Ratio Test. No modification of the expression is needed. 
A: We have
$$a_n = \dfrac{n!}{(2n-1)!!} = \dfrac{n!}{(2n)!} \times 2^n n! = \dfrac{2^n}{\dbinom{2n}n}$$
Use ratio test now to get that
$$\dfrac{a_{n+1}}{a_n} = \dfrac{2^{n+1}}{\dbinom{2n+2}{n+1}} \cdot \dfrac{\dbinom{2n}n}{2^n} = \dfrac{2(n+1)(n+1)}{(2n+2)(2n+1)} = \dfrac{n+1}{2n+1}$$
We can also use Stirling. From Stirling, we have
$$\dbinom{2n}n \sim \dfrac{4^n}{\sqrt{\pi n}}$$
Use this to conclude, about the convergence/divergence of the series.

EDIT
$$1 \times 3 \times 5 \times \cdots \times(2n-1) = \dfrac{\left( 1 \times 3 \times 5 \times \cdots \times(2n-1) \right) \times \left(2 \times 4 \times \cdots \times (2n)\right)}{ \left(2 \times 4 \times \cdots \times (2n)\right)}$$
Now note that
$$\left( 1 \times 3 \times 5 \times \cdots \times(2n-1) \right) \times \left(2 \times 4 \times \cdots \times (2n)\right) = (2n)!$$
and
$$\left(2 \times 4 \times \cdots \times (2n)\right) = 2^n \left(1 \times 2 \times \cdots \times n\right) = 2^n n!$$
Hence,
$$1 \times 3 \times 5 \times \cdots \times(2n-1) = \dfrac{(2n)!}{2^n \cdot n!}$$
A: Note that
$$
\frac{n!}{(2n-1)!!}=\frac11\frac23\frac35\frac47\cdots\frac{n}{2n-1}
$$
For $n\ge2$, $\dfrac{n}{2n-1}\le\dfrac23$. Thus, for $n\ge1$, we have
$$
\frac{n!}{(2n-1)!!}\le\left(\frac23\right)^{n-1}
$$
We can then compare with a geometric series.
