<s>[INST] Determine whether the following series converges or diverges I'm stuck on this problem:
$$1 + {1\cdot2\over1\cdot3} + {1\cdot2\cdot3\over1\cdot3\cdot5}+ {1\cdot2\cdot3\cdot4\over 1\cdot3\cdot5\cdot7} +\cdots$$
I've simplified the numerators $n!$ but can't figure out how to represent the denominators.
How do I go about solving this?
 [/INST] $$ 1 + {1\cdot2\over1\cdot 2\cdot3}2^1{(1)} + {1\cdot2\cdot3\over1\cdot2\cdot3\cdot4\cdot5}2^2(1\cdot2)+ {1\cdot2\cdot3\cdot4\over 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}2^3(1\cdot2\cdot3) +\cdots 
= \sum_{n=1}^{\infty}{n!(n-1)!2^{n-1} \over (2n-1)!}$$
It converges by ratio test
$$ \lim_{n\rightarrow \infty} {(n+1)!n!2^{n}\over (2n+1)!}\times {(2n-1)!\over n!(n-1)! 2^{n-1}} = \lim_{n\rightarrow \infty} {2(n+1)n  \over (2n+1)(2n)} = {1\over 2} < 1$$
For the value of convergence
$$1\cdot 3 \cdot 5 \cdots n=(2n-1)!!=\frac{2^n}{\sqrt{\pi}}\Gamma(n+\frac12)$$
The sum can be represented as 
$$ \sum_{n=1}^{\infty}{n! \over (2n-1)!!} = \sum_{n=1}^{\infty}{n! \sqrt \pi\over 2^n \Gamma(n + {1 \over 2})} = \sum_{n=1}^{\infty}{n! \sqrt \pi\over 2^n \left ( (2n)! \sqrt\pi \over n! 4^n\right )} = \sum_{n=1}^\infty {2^n (n!)^2 \over (2n)!}$$
We have the generating function
$$ \frac{ 4\,\left(\,{\sqrt{x+4}}-{\sqrt{x}}\,{\rm{arcsinh}}(\frac{{\sqrt{x}}}{2})\right) } { \sqrt{(x+4)^3} } =1-\frac{x}{2}+\frac{x^2}{6}-\frac{x^3}{20}+\frac{x^4}{70}-\frac{x^5}{252}+\frac{x^6}{924}-...$$
Choosing $x= -2$
$$ {4 \left ( \sqrt 2 + {\pi \over 2 \sqrt 2}\right ) \over 2 \sqrt 2} = 1 + \sum_{n=1}^\infty {2^n (n!)^2 \over (2n)!} $$
$$ \sum_{n=1}^\infty {2^n (n!)^2 \over (2n)!}  =  1 + {\pi \over 2}$$
Reference #1
Reference #2
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