Results 1 to 3 of 3

Math Help - Series Converge or Diverge

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    6

    Series Converge or Diverge

    A student was recently asked this question by his instructor:

    \sum_{n=1}^\infty{\left(\sqrt[n]{n}-1\right)}

    Converge or diverge?

    I feel a little dumb for not being able to answer it. The following tests fail to prove convergence or divergence:

    nth term test for divergence (limit is 0)
    ratio test (limit is 1)
    root test (see ratio test)
    limit comparison with \sqrt[n]{n} (not sure why I thought that'd work)

    Something I did try was using the fact that

    x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1}),

    to rewrite \sqrt[n]{n}-1 as

    \frac{n-1}{n^{1-1/n}+n^{1-2/n}+\cdots+n^{2/n}+n^{1/n}+1}.

    However, I'm not sure what to compare this to. According to wolfram alpha this series "diverges by the comparison test", but comparison to what? There is a similar problem in Baby Rudin, but for (\sqrt[n]{n}-1)^n, and a simple nth root test resolves that series [convergence] in a hurry. Any ideas? Have any of you encountered such an easy looking series before? Thanks.
    Last edited by fatpolomanjr; December 6th 2012 at 08:40 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Dec 2012
    From
    Danmark
    Posts
    11
    Thanks
    4

    Re: Series Converge or Diverge

    There you are, I hope it helps
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2011
    Posts
    6

    Re: Series Converge or Diverge

    Thank you for this!

    I found the answer through another recently, though. One way to show divergence was to use the inequality \ln(x)\leq n(x^\frac{1}{n}-1) for all x\geq1 and n\geq1. Then substituting n gives

    \frac{1}{n}\leq\frac{\ln(n)}{n}\leq \sqrt[n]{n}-1.

    Therefore the series diverges. Your solution is very nice because it gives a direct verification of the necessary inequality for the comparison test, which I find a bit more intuitive than "find the right Analysis inequality to get the job done" kind of argument.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Does this series converge or diverge?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 6th 2010, 09:15 AM
  2. Series: Converge of Diverge
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 6th 2010, 04:05 AM
  3. series - converge or diverge
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 10th 2009, 02:59 AM
  4. Does this series converge or diverge?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 30th 2009, 07:57 AM
  5. Series converge or diverge?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 1st 2008, 08:12 AM

Search Tags


/mathhelpforum @mathhelpforum