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Finding the sum of a series

  1. Jun 2, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the sum of the series e[tex]{(n + 1)/n)}[/tex] - e[tex]^{(n + 2)/(n + 1)}[/tex]

    3. The attempt at a solution

    Well for starters I got it into the proper format...IE

    (((n + 1)/(n))^n)/n! - (((n + 2)/(n + 1))^n)/n!

    But then I get a little lost...I would know how to take the limit, find convergence, divergence, but finding sums can be hard...I mean, it doesn't converge fast enough to just take the first few partial sums and be satisfied with that answer...help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 3, 2008 #2

    Dick

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    Have you considered the possibility that it's a cheap trick? Possibly a telescoping series?
     
  4. Jun 3, 2008 #3
    Very close!

    But, alas, putting numbers in for n for the first three terms gives...

    (2 - 1.5) + (1.125 - .88889) + (.39506 - .3255208)

    The powers are throwing it off...
     
  5. Jun 3, 2008 #4

    Dick

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    Uh, n=1. e^(2)-e^(3/2). n=2. e^(3/2)-e^(4/3). n=3. e^(4/3)-e^(5/4). What do you mean 'the powers are throwing it off'? You are the one who is 'throwing it off'.
     
  6. Jun 3, 2008 #5

    Dick

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    In the series expansion e^((n+1)/n)=sum(((n+1)/n)^k/k! over k). k and n aren't the same thing.
     
  7. Jun 3, 2008 #6

    Gib Z

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    Perhaps you would see what Dick is suggesting if you rewrote the summand as [tex]e ( e^{\frac{1}{n}} - e^{\frac{1}{n+1}})[/tex].
     
  8. Jun 3, 2008 #7
    Ohhh...I see.

    So, what is k supposed to be? Is it just a variable, or does it get a value?
     
  9. Jun 3, 2008 #8

    Dick

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    k is summed over in each exponential. They are both infinite sums. You've only selected the nth term from each sum. There are many more terms.
     
  10. Jun 3, 2008 #9
    Lemme see here...

    The series is indeed telescoping and I have it simplifying to...

    ((2^k)/k!) - (((n + 2)/(n + 1))^k)/k!

    Since n + 2 over n + 1 will run to 1...

    ((2^k/k!) - ((1^k/k!))

    So...

    1^k/k!

    Am I on the right track?
     
  11. Jun 3, 2008 #10

    Dick

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    No! That's not 'simplifying' it at all, except where it's wrong. You don't need to series expand the exponentials at all! The original series telescopes!
     
  12. Jun 3, 2008 #11

    HallsofIvy

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    There is a missing parenthesis here. Or do you mean
    [tex]e^{(n+1)/n}- e^{(n+2)/(n+1)}[/tex]?

    What do you mean by "proper format"?

    Actually write out the first few terms. An think about what Dick said.
     
    Last edited: Jun 3, 2008
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