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Homework Help: Sum of infinite geometric series

  1. Apr 10, 2008 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    I don't get what I am doing wrong here, I have attached my solution below. The solution manual have their answer as 3e/(3-e). Thanks!
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2008 #2

    Gib Z

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    Try posting the question in LaTex from now on, attachments have to be approved and put extra load on the moderators.
     
  4. Apr 10, 2008 #3
    I type really slow in Latex, I type faster in Mathtype in word, that's why I included as an attachment. Sorry. Anyhow, here is the link to that attachment

    http://p3t3rl1.googlepages.com/Eqn001.gif
     
  5. Apr 11, 2008 #4

    tiny-tim

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    Hi motornoob101!

    You got confused with your second āˆ‘.

    It should be eāˆ‘(e/3)^x-1, shouldn't it? :rolleyes:
     
  6. Apr 11, 2008 #5
    Erm I don't think get it.. So I have e^(x-1) which is same as (e^x) (e^-1) and e^-1 is (1/e) no?
     
  7. Apr 11, 2008 #6

    HallsofIvy

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    You have, skipping over your second term which simply makes no sense,
    [tex]\frac{e^n}{3^{n-1}}= \frac{1}{e}\left(\frac{e}{3}\right)^{n-1}[/tex]
    which is wrong- you have factored and e out of the numerator: it should be
    [tex]\frac{e^n}{3^{n-1}}= e\left(\frac{e}{3}\right)^{n-1}[/tex]
     
  8. Apr 11, 2008 #7
    Ohh I see. Thanks.
     
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