1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    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]


    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


    x-1 < 0

    x < 1

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


    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


    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]


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


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


    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook