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Math Help - show that the sum = sqrt(2)

  1. #1
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    show that the sum = sqrt(2)

    1+(1/4) + (1*3)/(4*8) + (1*3*5)/(4*8*12)+(1*3*5*7)/(4*8*12*16)+.......
    =sqrt(2)
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Nevermind! Listen to tonio.
    Last edited by Bruno J.; June 10th 2010 at 12:24 PM.
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  3. #3
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    Quote Originally Posted by bgbgbgbg View Post
    1+(1/4) + (1*3)/(4*8) + (1*3*5)/(4*8*12)+(1*3*5*7)/(4*8*12*16)+.......
    =sqrt(2)

    That's the sum 1+\frac{1}{4}+\sum^\infty_{n=2}\frac{1\cdot 3\cdot\ldots\cdots (2n-1)}{4\cdot (4\cdot 2)\cdot\ldots\cdot(4n)} . Let's try now to understand the quotient in that infinite series a little better:

    \frac{1\cdot 3\cdot\ldots\cdot (2n-1)}{4\cdot (4\cdot 2)\cdot\ldots\cdot (4n)}=\frac{1\cdot 2\cdot 3\cdot 4\cdot\ldots\cdot (2n)}{[4^n\cdot n!][2\cdot 4\cdot\ldots\cdot (2n)]} =\frac{(2n)!}{n!4^n\cdot n! 2^n} =\frac{(2n)!}{(n!)^22^{3n}} =\frac{(2n)!}{2^{2n}(n!)^2}\,\left(\frac{1}{\sqrt{  2}}\right)^{2n} . There's a reason why we wrote the last mess:

    We know (hopefully!) that \arcsin x=\sum^\infty_{n=0}\frac{(2n)!}{2^n(n!)^2(2n+1)}\,  x^{2n+1} , with convergence radius |x|<1 . Well, derivate both sides of this and get:

    \frac{1}{\sqrt{1-x^2}}=\sum^\infty_{n=0}\frac{(2n)!}{2^{2n}(n!)^2}\  ,x^{2n} , and now input x=\frac{1}{\sqrt{2}} , do a little algebra and get your result.

    Tonio

    Ps. This is, imho, a problem that fits the Challenge Problems Section!
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  4. #4
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    Another way to view the series is that it is the expansion by the Binomial Theorem of

    \left(1 - \frac{1}{2} \right)^{-1/2}.
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