1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Infinite series question

  1. Oct 13, 2005 #1
    Here's the question:
    Does the series SUM log(1+1/n) converge or diverge?

    I wrote out the nth partial sums like this:

    log(1+1) + log(1+1/2) + log(1+1/3)+..........+log(1+1/n)

    It looks to me like the limit of the thing inside the parentheses goes to 1 as n goes to infinity, making the limit of the entire thing 0. So I say it converges.

    One of my classmates says that SUM 1/n diverges, so this one does too. I can't disagree with him, but I fail to see the relevance.

    I'm confused. Any clarification will be appreciated.
  2. jcsd
  3. Oct 13, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    The terms going to zero is a necessary but not sufficient condition for convergence of the series (1/n is your basic counterexample).

    They probably have the limit comparison test in mind. You can also look at the integral comparison test, you can find an antiderivative of log(1+1/n) easily enough.
  4. Oct 13, 2005 #3
    The series diverges ..
    Take the sequance of partial sums ..
    [tex] S_n = \sum_{k=1} ^n\log \left( 1 + \frac 1k \right) = \sum_{k=1} ^n \log \left ( \frac { k+1} { k} \right )
    = \sum_{k=1} ^n \log (k+1) - \log (k) \mbox{ ( Telescoping series } [/tex]

    [tex] = (\log 2 - \log 1) + (\log 3 - \log 2) +....... + (\log n - \log (n-1) ) + (\log (n+1) - \log n) [/tex]

    [tex]= - \log 1 + \log ( n+1 ) = \log (n+1) [/tex]
    [tex] \lim _ { n \rightarrow \infty } S_n = \lim _ { n \rightarrow \infty } \log ( n+1) = \infty
    \Longrightarrow \sum_{n=1} ^ {\infty}\log \left( 1 + \frac 1n \right) \mbox{ diverges } [/tex]
    Last edited: Oct 13, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook