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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?

3 Answers 3

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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

  • 2
    More concisely: $\sum_{k=1}^\infty \frac{2^k}{\binom{2k}{k}}=\frac{\pi}{2}+1$2012-07-30
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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) $

  • 0
    You could phrase this answer like this: the series $S$ is bounded above, as every term is smaller than the corresponding term in the 3/4 geometric series. The 3.4 geometric series converges to a finite limit that gives an upper bound on the series $S$. The series $S$ is also increasing. Any increasing bounded sequence must converge.2012-07-30
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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.