Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

$$ 1 + {1\cdot2\over1\cdot 2\cdot3}2^1{(1)} + {1\cdot2\cdot3\over1\cdot2\cdot3\cdot4\cdot5}2^2(1\cdot2)+ {1\cdot2\cdot3\cdot4\over 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}2^3(1\cdot2\cdot3) +\cdots = \sum_{n=1}^{\infty}{n!(n-1)!2^{n-1} \over (2n-1)!}$$ 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

share|improve this answer
    
...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
2  
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 $$ S < a_1 + a_2 \left( 1 + \frac{3}{4} + \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 + \cdots \right) $$

share|improve this answer
    
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
1  
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
1  
@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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.