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Series: What do I have the right to do?

  1. Jan 15, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to say if the series converge or diverge.

    [tex]\infty[/tex]
    [tex]\sum[/tex] LN(n[tex]/2n+5[/tex])
    n=1

    2. Relevant equations

    I found this series to be divergent. However, I dont know if the way I proceeded is correct.


    3. The attempt at a solution

    I first separated the ln into two lns ( ln(n) - ln(2n+5 ). Then, I used it as a power of e to remove the lns, thus being left with -n -5. I then went for the limit divergence test, and it ended up diverging.

    Do I have the right to proceed that way?
     
  2. jcsd
  3. Jan 15, 2009 #2

    Dick

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    Are you summing ln(n/(2n+5))? If so, the first think you should check is whether the sequence ln(n/(2n+5)) approaches 0 as n->infinity. Does it?
     
  4. Jan 15, 2009 #3
    Well ...

    ln(n) - ln(2n+5) equals infinity minus infinity, which is an undetermined form. from there I'm just lost. I remembered about the e trick, but I'm not certain if I can use it in the case of series. This is probably very basic stuff but I havent done it in ages... I totally lost my "instinct". I dont want to waste anyone's time...
     
  5. Jan 15, 2009 #4

    Dick

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    Whats the limit of n/(2n+5) as n->infinity? Don't split it first.
     
  6. Jan 15, 2009 #5
    That would be infinity on infinity... If I use L'Hopital's rule, that would give me 1/2.
     
  7. Jan 15, 2009 #6
    You don't need to L'Hopitals. That's killing a cockroach with a nuke. Divide top and bottom by n.
     
  8. Jan 15, 2009 #7
    Hahaha. Alright. Resulting in 1/(2+5/n) and 5/n goes to zero.

    That gives us the limit of a constant, which would then be ln(1/2) ?
     
  9. Jan 15, 2009 #8

    AEM

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    Correct. And if you add up an infinite number of those ln(1/2)'s, what do you get?
     
  10. Jan 15, 2009 #9
    So since the limit does not equal zero, the series is divergent.
     
  11. Jan 15, 2009 #10

    Dick

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    That is so correct.
     
  12. Jan 16, 2009 #11
    Thanks everyone!
     
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