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



    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]

    email: danny.s.deng.ds@gmail.com
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