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Math Help - Root test for convergence/Divergence

  1. #1
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    Root test for convergence/Divergence

    sum 1-->infinity of (arctan(1/n)^n

    Here's what I've done:

    lim 1->infinity of: nth root (above expression)

    = lim 1-> infinity of arctan(1/n)

    Now what?
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  2. #2
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    Quote Originally Posted by jzellt View Post
    sum 1-->infinity of (arctan(1/n)^n

    Here's what I've done:

    lim 1->infinity of: nth root (above expression)

    = lim 1-> infinity of arctan(1/n)

    Now what?

    Well the properties of limits for composition functions, we can notice that the inside limit goes to 0 as we got to infinity, now we are approach the value of the arctan of 0 which is?

    Correct me if wrong.
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  3. #3
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    Any other advice on this one...
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  4. #4
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    Krizalid's Avatar
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    yes, brush up your notes from calculus I, since RockHard told ya what you need to do.
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  5. #5
    MHF Contributor chisigma's Avatar
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    At first the problem seems an easy application of a convergence criterion... may be however that it covers up an insidious trap ...

    From...

    http://en.wikipedia.org/wiki/Convergent_series

    Root test: suppose that the terms of the sequence are non negative and that there exists r such that...

    \lim_{n \rightarrow \infty} \sqrt[n] {a_{n}} = \frac{1}{r} (1)

    If r>1 the series converges, if r<1 the series diverges, if r=1 the root test is inconclusive...

    It is not quite obvious the reason to have in (1) \frac{1}{r} instead of r. Anyway in our case is...

    \lim_{n \rightarrow \infty} \sqrt[n] {a_{n}} = \lim_{n \rightarrow \infty} \tan^{-1} \frac{1}{n} = 0 (2)

    Now, if we want to apply the root test as explained by Wiky, the problem is that an r such that \frac{1}{r}=0 doesn't exists ...



    Merry Christmas from Italy
    Last edited by chisigma; December 10th 2009 at 06:32 AM.
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  6. #6
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    Well...I'd find wolfram to be more accurate than wiki. Wolfram states this converges by root test, if I entered the equation right
    Last edited by RockHard; December 10th 2009 at 05:30 PM.
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