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!

A_n := n^(1/n) - 1 yields a divergent series

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data
    I know that if [itex]a_n := n^{1/n} - 1[/itex], then [itex]\Sigma a_n[/itex] is divergent. I know this (by the integral test) because the integral of [itex]2^{1/n} - 1[/itex] from 1 to infinity is infinite. However, I want to avoid using non-elementary functions (here, the exponential integral) in my proof that this series is divergent.

    Can anyone see a way of doing this?


    2. Relevant equations/attempted solution
    [itex] lim\ sup_{n\to\infty} a_n^{1/n} = 1[/itex], so the root test is inconclusive. Comparison is getting me nowhere. I'm thinking about seeing whether using the Taylor expansion of each term of the sequence shows what I want to show...
     
  2. jcsd
  3. Aug 29, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I think the integral test is a good start. So we must prove that

    [itex]\int_1^{+\infty}{(2^{1/x}-1)dx}[/itex]

    diverges.

    Perhaps you can use the Taylor expansion of [itex]2^{1/x}[/itex] to find a good lower bound??
     
  4. Aug 29, 2011 #3
    Yes, thanks. As you'll see under number 2 above, that's exactly what I was considering doing.
     
  5. Apr 17, 2012 #4
    How about this:

    n^(1/n) - 1 > 2^(1/n) - 1;
    The definite integral of ( 2^(1/n) - 1 ) dn from 1 to infinity equals to
    the definite integral of (2^x - 1)(-1/(x^2)) dx from 1 to 0 by making a substitution (x=1/n),
    which, of course, equals to
    the definite integral of ( (2^x )/(x^2) - 1/(x^2) )dx from 0 to 1.
    Let f(x) = (2^x )/(x^2) - 1/(x^2), g(x)= - 1/(x^2) , then we have f(x) > g(x) for 0<x<1

    Since we know the definite integral of g(x) dx from 0 to 1 equals to infinity,
    we obtain that
    The definite integral n^(1/n) - 1 dn from 1 to infinity also equals to infinity. [Finished]


    Danny.
    email: danny.s.deng.ds@gmail.com
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook