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How can I simplify this?

  1. May 20, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to figure out the interval of convergence of the serie:


    [itex]\Sigma^{\infty}_{n=2} \frac{(-1)^{n}x^{n/2}}{nln(n)}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    The book states that the serie isn't a power serie, but a substitution can be used to transform it into a power serie. I tried to pose [itex]y^{n} = x^{n/2}[/itex].

    I also know that I must use the ratio test. Maybe there's another way to solve it, but the book states that the ratio test must be used. My problem is that I don't know how to simplify the ln(n+1) term that appears when I do that. Would anyone mind giving me a clue?
     
  2. jcsd
  3. May 20, 2012 #2

    sharks

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    Hi tamtam402

    Please use a descriptive title related to the actual problem in your post, according to the forum rules. This will help to bring more attention to your problem.
    [tex]\Sigma^{\infty}_{n=2} \frac{(-1)^{n}x^{n/2}}{nln(n)}[/tex]It appears that using the Alternative Series Test might be helpful as the first step.
     
  4. May 20, 2012 #3
    Unfortunately the book states that the problem must be done using the ratio test. I'm stuck with the ln(n+1) term, I don't know how to simplify it with the ln(n) term.
     
  5. May 20, 2012 #4

    sharks

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    What is x? Is it a constant? If yes, what value does it have? Is it greater than 0? You should provide all the information, or preferably the entire question. Type it all or just modify and attach a screenshot or picture to your first post.
     
  6. May 20, 2012 #5
    x isn't a constant, it's a variable since I'm working with a power serie. I wrote all the informations given in the book in my first post.

    The information given in the book is that I should do a substitution to transform the serie into a "real" power serie. Then I have to use the ratio test, and take the limit of n -> infinity on the ratio.

    According to a theorem, the power serie converges for the values of x where -1 < lim n-> infinity pn < 1

    The final step is to analyze the boundaries. However, I'm unable to simplify the ratio since I have a ln(n+1) term on the denominator and a ln(n) term on the numerator. I don't know how to simplify these.
     
  7. May 20, 2012 #6
    You cannot simplify algebraically ##\frac{\ln(n)}{\ln(n+1)}##. But there is a way to find ##\lim_{n\rightarrow\infty}\frac{\ln(n)}{\ln(n+1)}## using techniques from calculus. Hint: What are the limits of the numerator and denominator?

    And while it's not immediately applicable, the alternating series test might prove useful later in the problem.
     
  8. May 20, 2012 #7

    Ray Vickson

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    For small t > 0 we have ln(1+t) = t + O(t^2), so ln(n+1) = ln(n) + ln(1 + 1/n) = ln(n) + (1/n) + O(1/n^2) for large n. Using that should allow you to say enough about the ratio.

    RGV
     
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