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Math Help - limits

  1. #1
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    limits

    how do you solve something like this where there is a high exponent polynomial thing like

    lim_n->oo 2 / n^20(n+1)^20
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  2. #2
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    Quote Originally Posted by jeph View Post
    how do you solve something like this where there is a high exponent polynomial thing like

    lim_n->oo 2 / n^20(n+1)^20
    It is zero.
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  3. #3
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    ooh, ooh! I know this one!

    What you got here is a finite number basically divided by infinity^40 which is still just infinity. Any finite number divided by an inifinitly huge one is 0!
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  4. #4
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    Quote Originally Posted by Nerd View Post
    ooh, ooh! I know this one!

    What you got here is a finite number basically divided by infinity^40 which is still just infinity. Any finite number divided by an inifinitly huge one is 0!
    Or you can just say that,

    |2/n^20(n+1)^20| <= 1/n

    And lim n-->+oo (1/n)=0

    Then certainly that one goes to 0 as well.
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  5. #5
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    is that the direct comp test?

    the question was sumation n=1 oo 2^n / n^20. i used the ratio test to get 2/n^20(n+1)^20.

    on the sheet of paper i have it says 0<b_n<=a_n will diverge

    wouldnt 1/n be b_n in this case?

    1/n is bigger than 2/n^20(n+1)^20 so it wouldnt work...

    edit:
    oh wait...it = 0 so it converges....maybe i did something wrong then? because the series is supposed to diverge...

    can you show me how to solve sumation n=1 oo 2^n / n^20 to show that it diverges using the ratio test?
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  6. #6
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    Quote Originally Posted by jeph View Post
    is that the direct comp test?
    Almost! I used the "squeeze theorem" in disguise.

    Theorem: If {a_n} and {b_n} are convergent sequences to the same limit L and {c_n} is a sequence such that:
    a_n <= c_n <= b_n for sufficiently large n
    Then {c_n} is convergent to L as well.

    Theorem: If {a_n} is a convergent sequence to 0 and {c_n} is a sequence such that:
    |c_n|<=a_n for sufficiently large n
    Then {c_n} is convergent to 0.

    Proof: We can write |c_n|<=a_n differently:
    -a_n<=c_n<=a_n
    But the limit of a_n is 0 therefore the limit of -a_n is also 0.
    Hence by above theorem the limit c_n exists and equals 0.
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  7. #7
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    oh...i dont think we ever learned the squeeze thereom ._.
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