Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook