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Math Help - Sequence problem.

  1. #1
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    Sequence problem.

    For the following sequence, I'm supposed to plot it then guess what value it converges to (if it does). Then, I have to prove that my guess was right.

    #1: An = n^3/n!

    So, the first few terms of the seq are: 1,4,4.5,2.7,1.04,.3,etc....
    When graphed, it looks as if it might converge to zero.

    Now, I'm having a bit of trouble understanding the proof.

    The solution manual rearranges An to look like this:

    (n/n) * (n/(n-1)) * (n/(n-2)) *(n/(n-3))*....*(1/3)*(1/2)*(1)

    So far, this makes sense. But then they choose to pull out

    n^2/[(n-1)(n-2)(n-3)] for n>=4 and call it Cn, so that 0<An<=Cn. Then Squeeze Theorem can be used.

    So why did they choose n>=4? Why not n>=3 or n>=5? Is there another way to do this?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Hikari View Post
    For the following sequence, I'm supposed to plot it then guess what value it converges to (if it does). Then, I have to prove that my guess was right.

    #1: An = n^3/n!

    So, the first few terms of the seq are: 1,4,4.5,2.7,1.04,.3,etc....
    When graphed, it looks as if it might converge to zero.

    Now, I'm having a bit of trouble understanding the proof.

    The solution manual rearranges An to look like this:

    (n/n) * (n/(n-1)) * (n/(n-2)) *(n/(n-3))*....*(1/3)*(1/2)*(1)
    you have a typo here. the red n should be a 1 i think


    So far, this makes sense. But then they choose to pull out

    n^2/[(n-1)(n-2)(n-3)] for n>=4 and call it Cn, so that 0<An<=Cn. Then Squeeze Theorem can be used.

    So why did they choose n>=4? Why not n>=3 or n>=5? Is there another way to do this?
    lets say they said n>=3, what happens if n = 3? division by zero

    they chose the range of n so as to not cause problems with Cn, n <= 3, we have division by zero due to one of the factors (n - 1), (n - 2), or (n - 3)
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  3. #3
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    Oh, you're right about the typo. Sorry. ^^

    And thanks! I understand now.
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