1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    Hi motornoob101!

    You got confused with your second ∑.

    It should be e∑(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

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Sum of infinite geometric series
Loading...