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Homework Help: Series Convergence Test

  1. Mar 1, 2009 #1
    I am having problems with the following question:

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

    ([tex]\sum[/tex]nk=1 1/ekkx)n

    I used the ratio test to solve this but I'm not so sure about my solution:

    n1 = [tex]\frac{1}{e}[/tex]

    n2 = [tex]\frac{1}{e}[/tex] + [tex]\frac{1}{e^2 * 2^x}[/tex]

    n3 = [tex]\frac{1}{e}[/tex] + [tex]\frac{1}{e^2 * 2^x}[/tex] + [tex]\frac{1}{e^3 * 3^x}[/tex]

    n4 = [tex]\frac{1}{e}[/tex] + [tex]\frac{1}{e^2 * 2^x}[/tex] + [tex]\frac{1}{e^3 * 3^x}[/tex] + [tex]\frac{1}{e^4 * 4^x}[/tex]

    [tex]\sum all n[/tex] = [tex]\frac{n}{e}[/tex] + [tex]\frac{n-1}{e^2*2^x}[/tex] + [tex]\frac{n-2}{e^3*3^x}[/tex]

    So,

    Un = [tex]\sum[/tex]nn=1 [tex]\frac{n-(k-1)}{e^k*k^x}[/tex] = [tex]\frac{n-(k-1)}{e^n*n^x}[/tex]

    Un+1 = [tex]\sum[/tex]nk=1 [tex]\frac{(n+1)-(k-1)}{e^n*n^x}[/tex] = [tex]\frac{n-k+2}{e^(n+1) * (n+1)^x}[/tex]

    n-->[tex]\infty[/tex]

    Un+1/Un = [tex]\frac{(n-k+2)n^x}{(n-k+1)*(n+1)^x}[/tex]

    divide by n

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

    Lim[tex]\infty[/tex] nx-1 if convergent

    |nx-1| < 1

    so,

    x-1 < 0

    x < 1

    Does this look right?
     
  2. jcsd
  3. Mar 1, 2009 #2

    Mark44

    Staff: Mentor

    How did you get this?
    [tex]\sum all n = \frac{n}{e} + \frac{n - 1}{e^2 * 2^x} + \frac{n - 2}{e^3 * 3^x} [/tex]
     
  4. Mar 2, 2009 #3
    I assumed because the whole expression is encased in ()n that we summed all terms for each value of k upto n.
     
  5. Mar 2, 2009 #4

    HallsofIvy

    User Avatar
    Science Advisor

    I'm not sure why you are calling the "n1", "n2", etc. I presume you just mean "n= 1", "n= 2", etc.

    And, I'm not at all clear on what you are doing calculating all those terms. You said you used the ratio test but that's just
    [tex]\left(e^{-k}k^{-x}\right)\left(e^{k+1}(k+1)^{x}\right)= e\left(\frac{k+1}{k}\right)^k[/tex]
    What is that as k goes to infinity? For what x is it less than 1?
     
  6. Mar 2, 2009 #5
    Yes, I had misunderstood the ()n.

    I believe, effectively the series is :-

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

    If

    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]

    Therefore 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 :- ln term is negative,

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

    but is this a final solution?
     
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