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Ratio Test and Convergence

  1. Mar 3, 2009 #1
    I have been stuck on this question for a while and have not got much success from class mates...can somebody please help?

    Using an appropriate convergence test, find the values of x [tex]\in[/tex] R for which the following series is convergent:

    [tex]\sum[/tex]n=1n [tex]\frac{1}{e^n * n^x}[/tex]

    So,

    Un = [tex]\frac{1}{e^n * n^x}[/tex]

    and

    U(n+1) = [tex]\frac{1}{e^(n+1) * (n+1)^x}[/tex]

    then

    U(n+1)/Un = [tex]\frac{n^x}{e * (n+1)^x}[/tex]

    If the series is convergent:

    [tex]\frac{n^x}{(n+1)^x}[/tex] < e

    ln([tex]\frac{n^x}{(n+1)^x}[/tex]) < 1

    x * ln([tex]\frac{n}{n+1}[/tex]) < 1

    n/n+1 is always positive but less than one, so the Ln term is negative,

    So

    x > [tex]\frac{1}{ln(n/n+1)}[/tex]

    But the limit of this as n tends to infininty, is infinity, which doesnt make sense. Can I solve this for x, independent of n?
     
  2. jcsd
  3. Mar 3, 2009 #2
    Your trouble is in the last step. You divide both sides by the log without realizing that as n->+inf, it is equal to zero.

    Clearly, and you did all the work mind you, this is true for any x at all.

    Reason? |x ln(n/n+1)| < 1 for all x you can think of... since any real number times zero equals zero, and this is less than 1 in absolute value.

    Good job! Careful about that algebra, though...
     
  4. Mar 3, 2009 #3
    Thanks!!!
     
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