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Convergence of a Series

  1. Jun 11, 2005 #1
    For what values of [tex]x\in\mathbb{R}[/tex] does the series converge?

    [tex]\sum_{n=0}^{\infty}ne^{(-nx)}[/tex]

    I found this problem on the internet and couldn't solve it. Thanks for your help.
     
  2. jcsd
  3. Jun 11, 2005 #2

    saltydog

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    How about the integral test for convergence? For what values of x would the required integral converge?
     
  4. Jun 11, 2005 #3
    That is what I am having trouble with. I actually got this problem off of the MIT OpenCourseWare website under "Analysis I" and am looking to see how someone would solve this.

    When I evaluate [tex]\sum_{n=0}^{\infty}ne^{(-n)}[/tex] on my calculator, I get [tex]\frac{1}{4(\sinh{(\frac{1}{2})})^{2}}[/tex]. How was that calculated?
     
    Last edited: Jun 11, 2005
  5. Jun 11, 2005 #4

    shmoe

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    You could use the ratio test, though the integral test will also work. Have you tried to apply either?
     
  6. Jun 11, 2005 #5

    Zurtex

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    Well without getting too much in to detail of proving theorems to help with proving convergence. Think about what the sum would look like when x>0, when x=0 and when x<0.
     
  7. Jun 11, 2005 #6

    shmoe

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    It's related to a geometric series. Can you find

    [tex]\sum_{n=0}^{\infty}nx^n[/tex]

    (where it converges of course-actually if you can find where this guy converges then you should be able to convert back to your original problem without difficulty)
     
    Last edited: Jun 11, 2005
  8. Jun 11, 2005 #7
    I guess I can't. How would you do this? I know the [tex]\frac{1}{1-x}[/tex] answer for the value of an infinite series, but don't know how to apply it to this particular problem.

    I see that if I did find that value, let's say [tex]x=a[/tex], then I could replace [tex]x[/tex] with [tex]e^{(-x)}[/tex] and solve getting [tex]x=-\ln{(a)}[/tex].

    Would this be correct?
     
    Last edited: Jun 11, 2005
  9. Jun 11, 2005 #8
    Ok I tried this:

    [tex]\lim_{n\rightarrow\infty}\frac{|(n+1)x^{(n+1)}|}{|nx^{(n)}|}=x[/tex] so [tex]-1<x<1[/tex]. So now I replace [tex]x[/tex] with [tex]e^{(-x)}[/tex]. This is where I am stuck because I can't solve it. What did I do incorrectly?

    Thanks.
     
  10. Jun 11, 2005 #9
    hello all

    well to find out for what values this series converges for is not particularly hard its just a straight application of the ratio test the same we would apply it to the power series to find the radius of convergence

    [tex]\lim_{n\rightarrow\infty}\frac{(n+1)e^{(nx)}}{ ne^{(n+1)x}}=\frac{1}{e^x}<1[/tex]
    [tex]1<e^{x}[/tex]
    [tex]\log 1<\log e^{x}[/tex]
    and so therefore the series converges for all [tex]x>0[/tex]
     
  11. Jun 11, 2005 #10

    saltydog

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    Alex, you can do this.

    Let:

    [tex]w=\frac{1}{e^a}[/tex]

    and:

    [tex]z=\sum_{n=0}^{\infty}w^n[/tex]

    Now, you know what that sum is right? It's an expression in terms of w.

    Now take the derivative of both expressions with respect to w, adjust the sum to look like what you want and you're done.
     
  12. Jun 12, 2005 #11
    Doing that I come up with [tex]\sum_{n=0}^{\infty}nw^n=\frac{w}{(1-w)^2}\quad 0\leq w<1[/tex]. Is this correct?
     
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