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Homework Help: Convergence or divergence of log series

  1. Jul 11, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm supposed to evaluate the following series or show if it diverges:

    SUM (sigma) log [(x+1)/x]

    2. Relevant equations

    Drawing a blank...:confused:

    3. The attempt at a solution

    I'm unsure how to start this. We've gone over all sorts of tests for convergence (ratio, comparison, p-series, etc), but I'm not sure what to do with the log function. A push in the right direction would be appreciated!

    Thanks.
     
  2. jcsd
  3. Jul 11, 2007 #2

    CompuChip

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    What are you summing over (I assume x, though x usually denotes a continuous variables, and indices like k or n are commonly used for discrete summation) and what are your bounds (if x = 0 is included in the summation you immediately see that it diverges, but if [itex]x \ge 1[/itex] this is not immediately obvious).
    In other words, please be a little more specific.
     
  4. Jul 11, 2007 #3
    I recalled this incorrectly, it's log[(n+1)/n] and no bounds were given, though I would assume it's 1 to infinity as zero would diverge.
     
  5. Jul 11, 2007 #4
    ok n=1, n->infinity, before you think about the log, what happens to the (n+1)/n ? figure that out first then you can figure out if the log of that converges/diferges
     
  6. Jul 11, 2007 #5
    as n goes to infinity, (n+1)n will converge to 1, right? and log (1) is zero.
     
  7. Jul 11, 2007 #6
    Right, but that just tells you the series CAN converge, for this problem I'm thinking L'Hospital's rule due to the log.
    A nice series to compare with is the Harmonic Series, 1/x, since (n+1)/n = 1 +1/n.

    Edit: sry comparing with 1/x doesn't work >< Just woke up and automatically did L'Hospital's rule without checking if the comparison gave an indeterminate form.

    EditEdit: ok it does work >.> it gives you 0/0 lol
     
    Last edited: Jul 11, 2007
  8. Jul 11, 2007 #7

    quasar987

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    I suggest obtaining a lower bound on log [(x+1)/x] = log [1+1/x] by use of a Taylor expansion, and then comparing the sum of logs with the sum of that lower bound.
     
  9. Jul 11, 2007 #8

    NateTG

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    It's pretty easy to find a closed form for the partial sum.

    Big stinking giveaway hint:
    [tex]\log \frac{a}{b}=\log{a}-\log{b}[/tex]
     
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