1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Convergence or divergence of log series
Loading...