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Convergent/divergent with natural log

  1. Oct 11, 2009 #1
    Preparing for the math test, and cannot understand what to do for one of my practice problems:

    I need to find if the series is convergent or divergent, using test for divergence (divergent if limit doesn't equal 0)

    sigma from 1 to infinity of ln((n^2+1)/(2n^2+1))

    I see that I should use l'Hopital rule, and show that as n goes to infinity, the limit will equal to 1/2 (and series is diverges), but the ln (ie, natural log) gives me a doubt about my answer...

    Also, i figured that I can use one of the natural log properties and rewrite as sigma ln(n^2+1) - sigma ln(2n^2+1), but still i have no idea what to do with ln...

    help! what kind of influence does the natural log have on this problem??
  2. jcsd
  3. Oct 11, 2009 #2
    Another idea is that to do a l'Hopital rule on ln((n^2+1)/(2n^2+1)), which will flip my fraction, getting rid of natural log, and do l'Hopital rule 2 times more, and that will make my limit equals to 2, and show that the series is divergent...

    Is my logic solid, or does it have a hole somewhere?
  4. Oct 11, 2009 #3


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    There's no need for l'Hopital. It's easy to show that (n^2+1)/(2n^2+1) goes to 1/2 as n goes to infinity - just divide the numerator and denomonator by n^2. Since ln is continuous, the sequence converges to ln(1/2) =/= 0 so the series diverges.
  5. Oct 11, 2009 #4
    thank you very much
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