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Testing series with ln's for convergence

  1. Feb 12, 2007 #1
    I'm pretty embarassed that with a math degree I don't know how to do this. But today someone asked me to help him test this series for convergence:

    [tex]\Sigma \, ln(\dfrac{(k+1)^{2}}{k(k+2)} )[/tex]

    Maybe I'm just having a brain fart, but I can't think of any series to compare this to, and I obviously can't use the ratio, root, or integral tests. Any suggestions?
  2. jcsd
  3. Feb 12, 2007 #2
    One other question, if you guys don't mind. I saw the following series,

    [tex]\Sigma \, \dfrac{(k+2)}{2^{k}k(k+1)}[/tex]

    Anyway, it's pretty obvious that this series converges, but is there a way to precisely determine what it converges to?
  4. Feb 12, 2007 #3
    The first one isn't too hard using the rules of logs when you write them out in a series, realising you're actually considering the total multiplication of all the bits inside the ln( ... ) and then taking the ln.

    [tex]\ln \left( \frac{(k+1)^{2}}{k(k+2)} \right) + \ln \left( \frac{(k+2)^{2}}{(k+1)(k+3)} \right) + \ln \left( \frac{(k+3)^{2}}{(k+2)(k+4)} \right) = \ln \left( \frac{(k+1)(k+3)}{k(k+4)} \right) [/tex]

    Notice that the k+2 terms have cancelled along with one factor of the other numerators. Extending this you get

    [tex]\ln \left( \frac{(k+1)^{2}}{k(k+2)} \right) + \ldots + \ln \left( \frac{(k+n)^{2}}{(k+n-1)(k+n+1)} \right) = \ln \left( \frac{(k+1)(k+n)}{k(k+n+1)} \right) [/tex]

    Can you see from the 3 term case why it extends to this? All terms cancel except two in the first and two in the last.

    Split this up back into two expressions :

    [tex]\ln \left( \frac{(k+1)}{k} \right) + \ln \left( \frac{(k+n)}{(k+n+1)} \right)[/tex]

    Take n-> infinity gives the second term as ln(1) = 0. The first term depends on what you're taking k to be. If you start at k=1 then you end up with the total sum as ln(2).

    Job's a good 'un :)

    I'm not sure about the second one. I'd give it a try, but it's 2.40am and I should be asleep :eek:
  5. Feb 12, 2007 #4
    Wow, I can't believe I didn't think of that! Thanks a lot for your help.
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