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Radius and Interval of Convergence

  1. Dec 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the radius and interval of convergence for the two series:

    1) [(n+1)/n]^n * (x^n), series starting at n=1.

    2) ln(n)(x^n), series starting at n=1.


    2. Relevant equations
    You're usually supposed to root or ratio your way through these.


    3. The attempt at a solution

    1) First, combining the whole thing and putting to the nth power:
    [( (n+1) x) / n] ^ n

    then using the root test yields

    ((n+1)/n) * x

    but (n+1)/n doesn't converge. The book still gives R = 1 and I = -1 < x < 1, though... how?

    2) Uhhh not too sure... I just ratio tested it and got:

    (ln(n+1) / ln(n)) * x

    and the ln(n+1) / ln(n) doesn't converge. What do?

    Thanks.
     
  2. jcsd
  3. Dec 2, 2012 #2

    LCKurtz

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    Are you sure about that? Write it as$$
    1 +\frac 1 n$$
    For the second one think about L'Hospital's rule.
     
  4. Dec 4, 2012 #3
    Ok so giving this a second try, for the first one,
    using the root test, we end up with

    (1 + (1/n)) * x,

    so the limit is absolute value of x, which must be less than 1 to converge, so we set the equation

    -1< x< 1

    plugging -1 in to x we get , (-1)^n * (1+(1/n))^n , a diverging alternating series
    plugging 1 in to x we get (1)^n (1+ (1/n))^n, which diverges,

    so radius of convergence (R) = 1, and interval of con (I) = -1<x<1, open interval.

    For number 2,
    using LHospital's rule, limit is once again x, abs val of x is less than 1 for convergence so
    -1 < x < 1 for convergence,

    then plugging -1 in for x, we get (-1)^n * ln(n) which diverges,
    plugging in 1 for x, we get (1)^n * ln(n), which diverges,

    so R = 1, and I is the open interval -1 < x < 1.

    Amiriteuguise? thanks for any help.
     
  5. Dec 4, 2012 #4

    LCKurtz

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    It looks like you have the idea. Just remember that it is the ratio test or root test you are using, and when you say, for example, "then plugging -1 in for x, we get (-1)^n * ln(n) which diverges", what you really want to say is that (-1)^n * ln(n) doesn't converge to zero, hence the series ##\sum_{n=1}^\infty (-1)^n\ln n## diverges.

    And who/what is Amiriteuguise? Something to do with this thread?
     
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