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Test the series for convergence

  1. Apr 7, 2010 #1
    hello
    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

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

    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
    well,
    [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.
    [tex]
    1 - n^{\frac{1}{n}} = 2^k (1 - 2^{\frac{k}{2^k}})
    [/tex]

    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.
     
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