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Limit of series

  1. Feb 14, 2006 #1
    for infinity sum n=1 (2n+1) / (5n+1)

    how to find the converges or diverges??
    is it suitable to use basic comparison test??pls show me
  2. jcsd
  3. Feb 14, 2006 #2


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    If we split the fraction, we get

    [tex]\frac{{2n + 1}}{{5n + 1}} = \frac{2}{5} + \frac{3}{{25n + 5}}[/tex]

    As you can see, the second term will go to 0 if n tends to infinity but there's a remaining term 2/5. In other words: the limit of the associated sequence isn't 0 (for n going to infinity) and this is a necessary (though not sufficient) condition for the convergence of the series.
  4. Feb 14, 2006 #3


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    Or: divide both numerator and denominator by n:
    [tex]\frac{2n+1}{5n+1}= \frac{2+\frac{1}{n}}{5+\frac{1}{n}}[/tex]

    Now what does 1/n go to as n goes to infinity?
  5. Feb 14, 2006 #4
    but the question ask for diverges or converges??and the answer is diverges
  6. Feb 14, 2006 #5


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    When a series doesn't converge, it diverges.
  7. Feb 14, 2006 #6
    is it necessary to use any rules such as the one i mentioned??or just to perform the steps showed above??
  8. Feb 14, 2006 #7


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    I don't think those are necessary, if the limit of the (positive) sequence doesn't go to 0 (which is easy to check without using those other tests), then the (positive) series doesn't converge.
  9. Feb 14, 2006 #8
    teng the n-th term test for divergence is an easy check for this one.
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