# 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?

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$$1 + {1\cdot2\over1\cdot 2\cdot3}2^1{(1)} + {1\cdot2\cdot3\over1\cdot2\cdot3\cdot4\cdot5}2^2(1\cdot2)+ {1\cdot2\cdot3\cdot4\over 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}2^3(1\cdot2\cdot3) +\cdots = \sum_{n=1}^{\infty}{n!(n-1)!2^{n-1} \over (2n-1)!}$$ 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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...which diverges/converges to? And is this using the representation of the double factorial $1\cdot 3 \cdot 5 \cdots n=(2n-1)!!=\frac{2^n}{\sqrt{\pi}}\Gamma(n+\frac12)$? If so, where is the pi(e)? – draks ... Jul 30 '12 at 6:59
sorry ... i was verifying!! I have little experience with double factorial. – Santosh Linkha Jul 30 '12 at 7:01
strange: further down on the page, eq. (6), they give your formula. So it's left to you the show the con/divergence $\color{green}{.}\color{goldenrod}{.}\color{red}{.}$ – draks ... Jul 30 '12 at 7:05
to conclude multiply numerator and denominator by $n$ and use this – Raymond Manzoni Jul 30 '12 at 7:07
More concisely: $$\sum_{k=1}^\infty \frac{2^k}{\binom{2k}{k}}=\frac{\pi}{2}+1$$ – J. M. Jul 30 '12 at 10:25

The multipliers, like $4/7,$ are approaching $1/2,$ and in any case are below, say, $3/4$ as soon as you reach $3/5.$ So, with all positive summands, this compares favorably to a geometric series with ratio $3/4$ and converges.

BY AUDIENCE REQUEST: We are asking about $$a_1 + a_2 + a_3 + a_4 + \cdots,$$ where $a_1 = 1, a_2 = 2/3,$ then $a_3 < (3/4) a_2, a_4 < (3/4) a_3 < (3/4)^2 a_2,$ then $a_5 < (3/4)^3 a_2,$ and generally $a_n < (3/4)^{n-2} a_2.$ So any partial sum $S$ satisfies $$S < a_1 + a_2 \left( 1 + \frac{3}{4} + \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 + \cdots \right)$$

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Maybe its late.. but im having a hard time following your answer. – theone5526 Jul 30 '12 at 6:35
I cant see how the ratio is 3/4 ? – theone5526 Jul 30 '12 at 6:42
I think the simpler and more straightforward answer is to apply the ratio test. In fact, this answer just recapitulates the proof of the ratio test for a particular case. – Harald Hanche-Olsen Jul 30 '12 at 6:57
@haraldhanche-olsen Could you provide those steps in another answer please? The ratio test was my first intuition as well. – theone5526 Jul 30 '12 at 7:01
@theone5526: Not sure if I can find the time – but maybe later. It's standard textbook material, after all. – Harald Hanche-Olsen Jul 30 '12 at 7:19

Let us try to write down in a slightly simpler than above form the general term:

$$a_n:=\frac{1\cdot 2\cdot 3\cdot\ldots\cdot n}{1\cdot 3\cdot 5\cdot\ldots\cdot (2n-1)}=\frac{n!2\cdot 4\cdot 6\cdot\ldots\cdot 2n}{(2n)!}=\frac{2^n(n!)^2}{(2n)!}$$ and now just mimic what experiment did (D'Alembert's test):

$$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}[(n+1)!]^2}{(2n+2)!}\,\frac{(2n)!}{2^n(n!)^2}=\frac{2(n+1)^2}{(2n+1)(2n+2)}\xrightarrow [n\to\infty]{}\frac{2}{2\cdot 2}=\frac{1}{2}<1$$and the series converges.

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