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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.)

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This problem yells "ratio test" to me. And by the way, the numerator will also blow up. I hope you see that the bottom is (2n-1)! with half the terms missing. So twice as long with half the terms missing. What does your intuition say? – Fixed Point May 2 '13 at 5:14
I see now that $(2n-1)!!$ stands for odd numbered factorials. I did not know this until now, and I probably would have been lost for hours without this bit of knowledge! – Dylan May 2 '13 at 5:32
No worries ;-) that is how we all learn. I still lose hours because of what I don't know. – Fixed Point May 2 '13 at 6:01

Hint: Use the Ratio Test. No modification of the expression is needed.

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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.


$$1 \times 3 \times 5 \times \cdots \times(2n-1) = \dfrac{\left( 1 \times 3 \times 5 \times \cdots \times(2n-1) \right) \times \left(2 \times 4 \times \cdots \times (2n)\right)}{ \left(2 \times 4 \times \cdots \times (2n)\right)}$$ 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, $$1 \times 3 \times 5 \times \cdots \times(2n-1) = \dfrac{(2n)!}{2^n \cdot n!}$$

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Yikes. It is not at all apparent how $\frac{1}{(2n-1)!!}$ becomes $\frac{2^{n}n!}{(2n)!}$ – Dylan May 2 '13 at 5:38
Soon it will be obvious. Put in the "missing" terms $2,4,6,8,\dots,2n$. The product of what you just put in is $(2\cdot 1)(2\cdot 2)(2\cdot 3)\cdots(2\cdot n)$, which is $(2^n)(n!)$. – André Nicolas May 2 '13 at 5:48
Also, is the parenthesis thing a binomial coefficient? I am not sure how to solve those. Sounds like I need to go back and do more studying on these ones. – Dylan May 2 '13 at 5:50
I see now -- you have to divide out all the even numbered terms, which are represented by $(2^n)(n!)$. You guys make it sound as though it were so easy! Not so for people like me who suck at Math. Thanks. – Dylan May 2 '13 at 5:53

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.

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