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Limit of sum of exponential

  1. Jan 6, 2008 #1
    1. The problem statement, all variables and given/known data
    consider {sum from k=0 to n of e^(sqrt(k))}/{2sqrt(n)e^(sqrt)}.how to prove that the limit when n approaches infinity is 1?
    or in latex form,
    \lim_{n \to \infty}\frac{\sum_{k=0}^{n}e^{\sqrt{k}}}{2\sqrt{n}e^{\sqrt{n}}}=1

    2. Relevant equations
    Nil


    3. The attempt at a solution
    I tried to use logarithm to remove the exponential, but failed.
     
    Last edited: Jan 7, 2008
  2. jcsd
  3. Jan 7, 2008 #2

    Gib Z

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    Welcome to Physicsforums!

    That expression unfortunately is very hard to read in that form :( To get the LaTeX working on this forum, you must use the [ tex ] and [ /tex ] tags, without the spaces.

    I'll do that now, then try to help =]

    [tex]\lim_{n \to \infty}\frac{\sum_{v=0}^{n}e^{\sqrt{k}}}{2\sqrt{n} e{\sqrt{n}}}=1[/tex]

    EDIT: Ok it seems you meant;

    [tex]\lim_{n \to \infty}\frac{\sum_{k=0}^{n}e^{\sqrt{k}}}{2\sqrt{n} e^{\sqrt{n}}}=1[/tex]
     
  4. Jan 7, 2008 #3
    thanks for the help. i need the second equation, thanks for the edit :)
     
  5. Jan 7, 2008 #4
    it seems that the [tex]2\sqrt{n}[/tex] comes from the derivative of [tex]e^\sqrt{n}[/tex], but i couldn't think of anyway to make use of differentiation.
     
  6. Jan 7, 2008 #5

    Dick

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    Replace [tex]\sum_{k=0}^{n}e^{\sqrt{k}}[/tex] with [tex]\int_{0}^{n} e^{\sqrt{x}}dx[/tex] since it has the same behavior for large n (just like in the integral test for convergence). Now use l'Hopital.
     
  7. Jan 7, 2008 #6

    Defennder

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    Could you explain how this step is justified?
     
  8. Jan 7, 2008 #7

    CompuChip

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    Actually it seems to me that
    [tex]\sum_{k=0}^{n}e^{\sqrt{k}} - \int_{0}^{n} e^{\sqrt{x}}dx[/tex]
    is diverging :confused:
     
  9. Jan 7, 2008 #8

    Dick

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    Actually, I was hoping no one would ask for a detailed justification. Yes, the difference probably is divergent. But I think the denominator is even more divergent. One would have to show that the difference is small compared to the denominator. I'll try and come up with a good argument in a bit.
     
  10. Jan 7, 2008 #9

    Dick

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    Ok, try this. Apply l'Hopital first, using finite differences instead of derivatives. The difference between the nth partial sum and the (n-1)th partial sum in the numerator is e^(sqrt(n)). The difference between the value of the denominator at n and at n-1 is 2sqrt(n)e^sqrt(n)-2sqrt(n-1)e^sqrt(n-1). For large n I would approximate the denominator using a difference quotient for the derivative of 2sqrt(n)e^sqrt(n). There. That's the same thing, except I'm not making any claim that the integral 'approximates' the sum.
     
  11. Jan 7, 2008 #10

    Dick

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    Ok, now you're going ask how do I know the difference quotient can be approximated by a derivative. Skip that. Just directly show that the differences above approach the desired limit. It's the same machinery you'd use to show the difference quotient can be approximated by a derivative.
     
  12. Jan 7, 2008 #11
    guess you are using Stolz-Cesàro theorem. oh, now i need to prove that theorem? :rofl: thanks for the help, i will try to prove that Stolz-Cesàro theorem...
     
  13. Jan 7, 2008 #12
    if i apply what you said, i will need to prove
    [tex]\lim_{n \to \infty}\frac{e^{\sqrt{n}}}{2\sqrt{n}e^{\sqrt{n}}-2\sqrt{n-1}e^{{\sqrt{n-1}}}}=1[/tex]
    which gives
    [tex]\lim_{n \to \infty}\frac{1}{2\sqrt{n}-2\sqrt{n-1}e^{{\sqrt{n-1}-\sqrt{n}}}}=1[/tex]
    it then suffies to show that
    [tex]\lim_{n \to \infty}{2\sqrt{n}-2\sqrt{n-1}e^{{\sqrt{n-1}-\sqrt{n}}}}=1[/tex]
    i couldn't proceed from here. any hints?
     
    Last edited: Jan 7, 2008
  14. Jan 7, 2008 #13

    Dick

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    sqrt(n-1)-sqrt(n)=-1/(sqrt(n-1)+sqrt(n)). Expand the exponential to first order. Oh, yeah, Stolz-Cesaro, that one. I figured it must have a name, but I didn't know it.
     
  15. Jan 7, 2008 #14
    I got it. A million thanks to those who helped. :smile:
     
  16. Jan 8, 2008 #15

    Gib Z

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