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Does SUM log(1+1/n)/n converge?

  1. Mar 15, 2006 #1
    Prove that [tex]\sum \frac{1}{n}\log(1+\frac{1}{n})[/tex] converges.

    I tried a batch of tests, but none works. Last time Raabe's test was give to me upside down. I wonder if any other tests are give to me wrongly as well.
     
  2. jcsd
  3. Mar 15, 2006 #2
    Have you tried comparing it to 1/n^2?
     
  4. Mar 15, 2006 #3
    I can't get it to work out
     
  5. Mar 15, 2006 #4

    shmoe

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    Post the details of your comparison...

    What bounds do you know for log(1+x)?
     
  6. Mar 15, 2006 #5
    It's unbounded.

    I figured if [tex]\exp(\frac{1}{n}\log(1+\frac{1}{n}))=\exp(\frac{1}{n})+\exp(\frac{1}{n})/n \rightarrow 1[/tex] as [tex]n\rightarrow\infty[/tex], so at least the terms go to zero, although it's insufficient.

    I could use the Mercator series, but we haven't seen it yet.
     
    Last edited: Mar 15, 2006
  7. Mar 16, 2006 #6

    HallsofIvy

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    The first thing that was suggested was comparison- have you tried that?

    The Taylor's series for ex is 1+ x + (1/2)x2+ ... so if x is positive, ex> 1+ x and x> log(1+ x) for all x. In particular, 1/n> log(1+ 1/n) for all n.
     
  8. Mar 16, 2006 #7

    shmoe

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    It's bounded by a constant in the region you're concerned with. You are looking at log(1+x) where x=1/n, so 1<1+x<2.

    A constant bound will not suffice here though, you need some indication of how fast this term is going to zero. Try the log(1+x)<x Halls mentioned, this was the one I was fishing for (one function can "bound" another).
     
  9. Mar 16, 2006 #8
    I got it. Thanks for the help.
     
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