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Infinite series problem

  1. Oct 5, 2010 #1
    1. The problem statement, all variables and given/known data
    \sum_{k=2}^\infty \frac{k^2}{k^2-1}

    \sum_{n=1}^\infty \frac{1+2^n}{3^n}

    2. Relevant equations

    I know that for the first problem i can apply the Divergence test by finding my limit as K goes to infinity. By doing this i get 1 which does not equal zero so i know it diverges.

    Now my question is why cant i apply this same test to the 2nd problem? it seems as n approaches infinity i would get 1 as well. but thats not the case the correct solution for the 2nd problem is as follows..

    \sum_{n=1}^\infty \frac{1}{3^n} + \frac{2^n}{3^n}

    Then by using a little algebra and sum of two convergent geometric series we get 5/2 to be the answer. which is convergent.

    So why cant i find my limit as n approaches infinity in the 2nd problem? and why is 5/2 convergent? i thought if r>1 the series diverges?
    Any help is appreciated.
  2. jcsd
  3. Oct 5, 2010 #2
    You can apply the Divergence Test to the 2nd problem, but it doesn't tell you anything because


    so the series may or may not diverge (obviously you already know that it converges, but in the future..).
  4. Oct 5, 2010 #3


    Staff: Mentor

    The series is convergent because it is the sum of two convergent geometric series.
    It makes no sense to say that "5/2 is convergent." For the first series, r = 1/3, making it a convergent geometric series. For the second series, r = 2/3, making it also a convergent geometric series.
  5. Oct 6, 2010 #4
    Awesome. Thank you both for the help. I missed out on a this lecture so i was trying to teach it to myself by using the text and was clearly having a hard time. Everything seems to make much better sense now
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