Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Test the series for convergence

  1. Apr 7, 2010 #1
    I have this one:

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

    mmmmm am sure it will be tested by using one of the comparison tests
    but am not getting it
    any help?

    this is not my homework, actually I finished my college 2 years ago.
  2. jcsd
  3. Apr 7, 2010 #2
    I would start by looking at the behavior of just [itex]\sqrt[n]{n}[/itex] as [itex]n\to\infty[/itex]. Use the techniques of logarithms to find [itex]\lim \sqrt[n]{n}[/itex]. Once you see what [itex]\sqrt[n]{n}[/itex] approaches, you should immediately be able to see how

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

    must behave. If you don't see it, just consider what [itex](1-\sqrt[n]{n})[/itex] must approach, knowing the limit of [itex]\sqrt[n]{n}[/itex].
  4. Apr 7, 2010 #3
    [tex]1 - \sqrt[n]{n} \rightarrow 0[/tex] as [tex]n \rightarrow \infty[/tex]

    and this will make no sense.
  5. Apr 7, 2010 #4
    You are quite right; in my zeal, I made a mistake in computing the limit of the n-th root of n and getting 0!
  6. Apr 7, 2010 #5
    I might be wrong but try Cauchy's Condensation Test.
    1 - n^{\frac{1}{n}} = 2^k (1 - 2^{\frac{k}{2^k}})

    Which obviously fails the limit test...
  7. Apr 8, 2010 #6
    It must be solved by the standart test.

    Anyone ?
  8. Apr 8, 2010 #7
    In order to use the CCT your terms need to be positive and non-increasing. This isn't a big deal since we can just negate the sum, and consider the sum starting from n=3.

    As for using "standard" tests, what about the integral test? I've only thought as far as:

    [tex]\int_3^\infty (\sqrt[x]{x}-1)\ dx \ge \int_3^\infty (\sqrt[x]{3}-1)\ dx = \int_3^\infty (3^{1/x}-1)\ dx[/tex]
  9. Apr 9, 2010 #8

    Gib Z

    User Avatar
    Homework Helper

    Theorem: A bounded monotonic sequence converges.
  10. Apr 9, 2010 #9
    Yes, but a sequence is very different from a series. Unless you are referring to the partial sums; but this would require that you can bound the partial sums. Can you elaborate on how this is done?
  11. Apr 9, 2010 #10

    Gib Z

    User Avatar
    Homework Helper

    The n-th root of n is greater than the n-th root of 1.
  12. Apr 9, 2010 #11
    Still searching for a solution with standard tests ..
  13. Apr 9, 2010 #12
    Try finding a function f(x) such that [tex]\sqrt[x]{x} - 1[/tex] >> f(x) using l'Hôpital's rule, [tex]\sqrt[x]{x} - 1[/tex] > f(x) on [1, ∞), and [itex]\sum_{n=1}^\infty f(n)[/itex] diverges. Then use the comparison test on your series and f(n).
    Last edited: Apr 9, 2010
  14. Apr 12, 2010 #13
    I tried that
    but its not easy to find that f
    and also f must be positive
  15. Apr 12, 2010 #14
    [itex]n^{1/n}-1[/itex] is asymptotic to [itex]\log(n)/n[/itex], so you need to analyze the convergence of [itex]\sum\log(n)/n[/itex]
  16. Apr 12, 2010 #15
    Ahh, this indeed will do it. Very clever.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook