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Series sums

  1. Aug 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the sum of the series: [tex]\sum^{\infty}_{0}[/tex](e[tex]^{\frac{n+1}{n}}-e^{\frac{n+2}{n+1}}[/tex])


    2. Relevant equations
    Telescoping series equation.

    3. The attempt at a solution
    Starting with n=0, I get: (e[tex]^{2}[/tex]-e[tex]^{\frac{3}{2}}[/tex])+(e[tex]^{\frac{3}{2}}-e^{\frac{4}{3}})+(e^{\frac{n+1}{n}}-e^{\frac{n+2}{n+1}}[/tex])=e[tex]^{2}[/tex] as n approaches infinity. Is this correct?
     
  2. jcsd
  3. Aug 10, 2008 #2
    (e^2-e^4/3)
    +
    (e^4/3 - e^5/4)
    +
    (e^5/4 - e^6/5)
    +
    ...
    +e^(n+1/n - e^(n+2)/(n+1))

    =
    e^2 - e^(n+2)/(n+1) <-- you forgot this.

    And should it be starting with n =1?
     
  4. Aug 10, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that's a "telescoping" series- every term except the first cancels a later term. Well done.
     
  5. Aug 10, 2008 #4
    Alright! Thanks!
     
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