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Infinite series, does ∑ n/2^n diverge?

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\sum^{\infty}_{n=1} \frac{n}{2^{n}}[/tex]

    Does this series converge or diverge?

    2. Relevant equations

    3. The attempt at a solution

    By the Cauchy condensation test (http://en.wikipedia.org/wiki/Cauchy_condensation_test) I think this one diverges. But not sure if I am using it correctly.

    According to the test,

    [tex]\sum^{\infty}_{n=1} \frac{n}{2^{n}}[/tex]

    converges if and only if

    [tex]\sum^{\infty}_{n=1} 2^{n} \frac{2^{n}}{2^{n}} = \sum^{\infty}_{n=1} 2^{n}[/tex]

    converges, which doesn't.

    Thank you for any help.
  2. jcsd
  3. Oct 20, 2009 #2


    Staff: Mentor

    In the line above, you aren't using the condensation test correctly. For your series, f(n) = n/(2n), so what would be f(2n)?

    A test that would be simpler to apply would be the Limit Ratio Test.
  4. Oct 20, 2009 #3
    Ah, okay, I see it where I went wrong..

    Using the limit ratio test

    [tex] lim \left| a_{n+1}/a_{n} \right| = lim \left| \frac{(n+1)/2^{n+1}}{n/2^{n}} \right| = 1/2 < 1 [/tex]

    So it converges...
  5. Oct 21, 2009 #4


    Staff: Mentor

    Yes, indeed.
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