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McLauren Series

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data

    The MacLaurin series for ln(1/1-x) is sum of x^n/n with interval of convergence -1<= x <1

    a. Find the Maclaurin series for ln(1/1+3x) and determine the interval of convergence
    b. Find the value of [tex]\sum[/tex] (-1)^n/n from n=1 to infinity.
    c. Give a value of p such that [tex]\sum[/tex] (-1)^n/ n^p converges but [tex]\sum[/tex] of 1/n^(2p) diverges. Give reasons why your value of p is correct.
    d. Give a value of p such that [tex]\sum[/tex]1/n^p diverges but [tex]\sum[/tex]1/n^(2p) converges. Give reasons why your value of p is correct.

    2. Relevant equations

    Limit test

    3. The attempt at a solution

    a. So I guess the answer for the series is [tex]\sum[/tex](-1)3x^n/n. Am I right ?
    b. How do I do this one ?
    c. I am not sure how to do this one.
    d. I think 1/n^p diverges when p<1. For p<1, 2p<2 so 1/n^2p < 1/n^2 which is a convergent series. So my p is correct. Am I right ?
  2. jcsd
  3. Mar 31, 2009 #2
    I think I did part d wrong to. Right ?
  4. Apr 1, 2009 #3


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    Science Advisor

    No, you are not. HOW did you "guess the answer"?

    Try something!

    You are not doing the right problem. You were asked to find p such that [itex]\sum (-1)^n/n^p[/itex] converges and [itex]\sum 1/n^p[/itex] diverges.
  5. Apr 1, 2009 #4
    a. I actually did not guess, I multiplied x by 3 and negated it. If am wrong, how should I do it then ?

    d. Oh, I see. I misunderstood the problem. Then p should be 1 then, right ? since 1/n diverges but 1/n^2 converges
  6. Apr 1, 2009 #5


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    Homework Helper

    just to conceptualise it, try re-writing, first from your problem description you have (i haven't checked this...):
    [tex]ln(\frac{1}{1-x}) = \sum \frac{x^n}{n}[/tex]

    and you want to find
    [tex]ln(\frac{1}{1+3y}) = [/tex]

    can you find x = f(y) and substitute this into you original taylor expansion?

    d) sounds reasonable if p is an integer, can you expain why though? think convergence tests
  7. Apr 1, 2009 #6
    . How should I do part b ?
  8. Apr 1, 2009 #7
    ln(1/1+3x) = ln(1/1-(-3x))

    You had ln(1/1-x) and now you are multiplying -3 to x
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