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Math Help - Series and sequence problem

  1. #1
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    Question Series and sequence problem

    Some one help please. Have been struggling with this for more than an hour..

    sum (n/(n+1))^(n^2), where n goes from 1 to infinity.

    How to show that this series converge when n goes from 1 to infinity?
    I try root test but L goes to one, which mean the test is inconclusive (thus ratio test doesn't work, too).

    Thanks so much!
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  2. #2
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    Do you recognize the indeterminate form, 1^{\infty}
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Do you recognize the indeterminate form, 1^{\infty}
    Do you mean (n/(n+1))^(n^2) is equivalent to 1^(infinity)? I understand that when n approaches infinity n/(n+1)=1, so this limit doesn't go to 0 the series will be divergent?
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  4. #4
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    No, my question was particularly about the "Indeterminate Form". If we recognize that, we can employ l'Hospital! What say you? Shall it go that way or otherwise?
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  5. #5
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    Okay. I see what you mean. But as far as I've learned I only use L'Hospital on things like 0/0 infinity/infinity. Does it have to be in a fraction form?
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  6. #6
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    Consider the logarithm of the general term.
    Use log rules to get the n^2 out of the exponent.
    Rewrite the n^2 as an awkward reciprocal in the denominator.

    Think about this as an Indeterminate Form that you are used to seeing.

    Use log rules to simplify the numerator
    Apply l'Hospital.

    See if we are getting anywhere.
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  7. #7
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    I've tried that before but I am stuck at (n^2)ln(n/(n+1)). I changed it to (n^2)/(1/ln(n/(n+1))) and found that it was infinity/1 form.
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  8. #8
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    Well, I suggested putting 1/n^2 in the denominator, but this will do for the demonstration.

    It's not "1" in the denominator. log(1) = 0 ==> And the reciprocal of that would be...

    Now, put it back in the numerator and put the 1/n^2 in the denominator. Trust me on this. Your life will be much easier.
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  9. #9
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    But still we can't tell if this series is convergent even if we get the limit = 0. As I said before I am supposed to show this series is convergent.
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  10. #10
    Ted
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    Use the Root Test.
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  11. #11
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    Quote Originally Posted by Ted View Post
    Use the Root Test.
    cjt422 said in his first post that he had tried the root test and it was not conclusive (the limit was 1).
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  12. #12
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    Sure you can. You have an integral test, don't you? Can the integral fail to converge if the limit of the term is zero?
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