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Homework Help: Convergence: Root Test Inconclusive

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm this series to see if it's convergent or divergent. I tried the root test, but it came out inclusive, and now I am trying to figure out if the ration test works. The only thing I'm asking which would be the right test for this.

    2. Relevant equations

    ∞Ʃn=1 (2n+1)^n/(n^2n)

    3. The attempt at a solution

    = (2n+1)/(n^2)
    = [itex]\stackrel{}{lim n\rightarrow}[/itex]∞ (2n+1)/(n^2)
    = 1/1
    = 1
  2. jcsd
  3. Nov 17, 2011 #2


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    Oh, come on. The limit of (2n+1)/n^2 is not 1.
  4. Nov 17, 2011 #3
    Oh, jeez! You're right! I made a dumb mistake and forgot to divide the 1 in (2n+1) by n^2. Oh, and thanks.
  5. Apr 12, 2012 #4
    hello, i have another problem. Could you help me ?
    The problem is ;

    ∞Ʃn=0 (1+1/n)^n

    When I use root theorem;

    n√((1+1/n)^n) = (1+1/n)

    lim n→∞ (1+1/n) = 1

    Result is inconclusive so what should I do ?

    thanks in advance
  6. Apr 12, 2012 #5


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    Try the nth term test. http://en.wikipedia.org/wiki/Term_test
  7. Apr 13, 2012 #6
    Ok I tried it:

    lim n→∞ (1+(1/n))^n = e (it is written in some theorem)

    And rule of nth term test is
    If lim n→∞ an ≠0 or if the limit does not exist, then ∞Ʃn=1 an diverges.

    My result is 'e' is not equal 0 so, ∞Ʃn=1 an diverges.

    My original question :

    ∞Ʃn=0 an
    Is there any problem that in my question the sigma starts with 0 but nth term test rule says sigma starts with 1.

    Does it change anything ?

    If not, my result is divergence.
  8. Apr 13, 2012 #7


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    No, it doesn't change anything, and that's an important point. All that matters for convergence or divergence is what happens 'near infinity'. Changing any finite number of terms of your series doesn't affect convergence.
  9. Apr 13, 2012 #8
    Ok, I got it. Thank you so much.
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