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Another Question Regarding Power Series

  1. Nov 1, 2004 #1
    I used Mathematica to confirm that

    [tex] \sum _{n=0} ^{\infty} \left[ \frac{\left( -1 \right) ^n \left( 2n \right) x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right] + \sum _{n=0} ^{\infty} \left[ \frac{\left( -1 \right) ^n \left( 2n \right) \left( 2n+1 \right) x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right] = - \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n x^{2n+3}}{n! \left( n+1 \right) ! 2^{2n+1}} [/tex]

    Does anyone know how to obtain it directly?

    Thank you.
     
  2. jcsd
  3. Nov 1, 2004 #2

    shmoe

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    Science Advisor
    Homework Helper

    Sure, combine the series on the left and notice that when n=0, the coefficient vanishes to get:


    [tex] \sum _{n=1} ^{\infty} \left[ \frac{\left( -1 \right) ^n \left( 2n \right) (2n+2) x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right][/tex]

    Factor the numerator slightly:

    [tex] \sum _{n=1} ^{\infty} \left[ \frac{\left( -1 \right) ^n \left( n \right) (n+1)2^2 x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right][/tex]

    Do some light cancelling, write our power of x in a funny way and lets pull out a -1 for fun:

    [tex]-\sum _{n=1} ^{\infty} \left[ \frac{\left( -1 \right) ^{n-1} x^{2(n-1)+3}}{(n-1)! \left( (n-1)+1 \right) ! 2^{2(n-1)+1}} \right][/tex]

    You are now a change of index variable away from glory.
     
    Last edited: Nov 1, 2004
  4. Nov 1, 2004 #3
    Thanks a lot! It makes sense now.
     
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