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Series convergence

  1. Nov 29, 2013 #1
    1. The problem statement, all variables and given/known data

    An=Ʃ(k)/[(n^2)+k]
    the sum is k=0 to n, the question is, to which value does the this series converge to
    2. Relevant equations
    i know for sure that this series converges, but could not figure out the value to whch it converges


    3. The attempt at a solution

    i did the convergence test, mod(An+1/An).. the value is 1. what i do now?
     
  2. jcsd
  3. Nov 29, 2013 #2

    Dick

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    You want to use the squeeze theorem. Try and bracket that sum between two sums whose limit you can evaluate and whose limit turn out to be the same. Here's a hint. The k in the denominator is what's making it hard to evaluate.
     
  4. Nov 29, 2013 #3
    actaully i don't want to find the limit of the fn. i want the sum of the series.
     
  5. Nov 29, 2013 #4

    Dick

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    'converges' means you take the limit of the sums ##A_n## as n->infinity. There's no simple closed form expression for the sum. There is a simple expression for the limit of the sum. This isn't really the same as the usual infinite sum problem. ##A_n## isn't the partial sum of some series. The upper limit n is in the expression for the terms you are summing.
     
    Last edited: Nov 29, 2013
  6. Nov 29, 2013 #5

    Ray Vickson

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    Using the "test"
    [tex] \left|A_n + \frac{1}{A_n} \right| [/tex]
    (which is what you WROTE) will get you nowhere. Even the correct test ##A_{n+1}/A_n## (written as A_{n+1}/A_n or A_(n+1)/A_n) will still get you nowhere: your problem is NOT to decide on convergence of an infinite series.

    The summation cannot be performed in elementary terms: Maple gets the answer
    [tex] \sum_{k=0}^n \frac{k}{n^2+k} = n+1+n^2 \Psi(n^2)-n^2 \Psi(n^2+n+1)[/tex]
    where ##\Psi(x)## is the so-called "Psi function" or "di-Gamma function", defined as the logarithmic derivative of the Gamma function ##\Gamma(x)##:
    [tex] \Psi(x) \equiv \frac{d \ln(\Gamma(x))}{dx} = \frac{\Gamma\, ^{\prime} (x)}{\Gamma(x)}.[/tex]
     
    Last edited: Nov 29, 2013
  7. Nov 30, 2013 #6
    @ray wickson,
    could you tell me an answer for this.
     
  8. Nov 30, 2013 #7

    Dick

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    Nobody is going to 'tell you an answer', you have to work on it. You really don't need any nonelementary functions for this. Just think about comparison series.
     
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