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Convergence and divergence of a series

  1. Oct 10, 2014 #1
    B]1. The problem statement, all variables and given/known data[/B]
    Find whether the series is convergent or divergent

    2. Relevant equations

    lim.JPG

    3. The attempt at a solution

    By ratio test I have,
    limit.JPG

    I would apply L'Hôpital's rule to find the value of limit but before that how do i simplify the expression? It has fractional part both in the numerator as well as in the denominator.
     
    Last edited: Oct 10, 2014
  2. jcsd
  3. Oct 10, 2014 #2

    Ray Vickson

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    Just use elementary high-school algebra. Alternatively, look more carefully at the problem before even starting. Maybe the ratio test won't work; there are times when it doesn't.
     
  4. Oct 10, 2014 #3

    RUber

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    Consider breaking it into two sums...the sum of convergent series is convergent, however if one diverges, the sum of the two diverges (generally).
     
  5. Oct 10, 2014 #4

    statdad

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    If you want to simplify the large fraction in
    [tex]
    \lim_{n\to \infty} \left(\dfrac{\left(\dfrac{(n+1)^2}{2^{n+1}} + \dfrac{1}{(n+1)^2}\right)}{\dfrac{n^2}{2^n} + \dfrac 1 {n^2}}\right)
    [/tex]

    treat it the way you would a complex fraction. As has been stated above, however, I'm not sure this approach will generate a positive result.

    Think about the idea that if both [itex] \sum_{i=1}^\infty a_n[/itex] and [itex] \sum_{i=1}^\infty b_n [/itex] are absolutely convergent then
    [itex] \sum_{i=1}^\infty \left(a_n + b_n \right) [/itex] is absolutely convergent.
     
  6. Oct 10, 2014 #5

    LCKurtz

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    Not sure what you mean by "generally" other than perhaps it means "sometimes" because you know it's false in general.
     
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