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Simplify the series

  1. Jul 31, 2012 #1
    1. The problem statement, all variables and given/known data
    This is a sub problem, the whole problem is find interval of convergence and radius of convergence but the thing im having trouble with is how to simplify this
    simplify the series

    2*4*6*8*(2n)/2*4*6*8....(2n+2)

    2. Relevant equations



    3. The attempt at a solution

    keys say 2*4*6*8...(2n)=2^n*n!

    so 2^n*n!/2^(n+1)*(n+1)!=1/(2n+2)

    my question is, HOW??? how do u go from 2*4*6*8...(2n) to 2^n*n! or how do u get from

    2*4*6*8*(2n)/2*4*6*8....(2n+2) to 1/(2n+2)

    sorry for not using the proper tools, they are very confusing :(
     
  2. jcsd
  3. Jul 31, 2012 #2

    SammyS

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    I'll address, "HOW??? how do u go from 2*4*6*8...(2n) to 2^n*n! ?" .

    [itex]2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\dots(2n-2)\cdot(2n)[/itex]
    [itex]=(2\cdot1)(2\cdot2)(2\cdot3)(2\cdot4)(2\cdot5)(2 \cdot6)(2\cdot7)\dots2(n-1)\cdot2(n)[/itex]

    [itex]=2^n\cdot1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7 \dots(n-1)\cdot(n)[/itex]

    ...​
     
  4. Jul 31, 2012 #3

    Dick

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    For example, 2*4*6*8=(2*1)*(2*2)*(2*3)*(2*4)=(2*2*2*2)*(1*2*3*4)=2^4*4!. Just think about it a little. It's not magic.
     
  5. Jul 31, 2012 #4

    LCKurtz

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    2*4*6*8...*(2n) = (1*2)*(2*2)*(3*2)*(4*2)...(n*2) The first factor in each parentheses gives the n! and the second gives 2n.
    In the denominator the factors increase by 2 each factor. What would the factor right in front of the (2n+2) be?

    [Edit]Guess I have to learn to type faster. While I'm typing, two other responses appear.
     
  6. Aug 2, 2012 #5
    thanks for replies! appreciate it
     
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