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Math Help - Does this sequence converge?

  1. #1
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    Does this sequence converge?

    Is this correct?

    The dominant term is n!, so we write

    {a}_{n } = \frac{{n}^{4 }+2(n!) }{{n}^{3 }+{(-1)}^{n+1 }(n!) } = \frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } }

    Since { {n}^{4 }/n! },{ {n}^{3 }/n! } and { {(-1)}^{n+1 } } are all null sequences,

    \lim_{n \to \infty} {a}_{n } = \lim_{n \to \infty} \frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } } = \frac{0+2}{0+0 } = 0

    by the Combination Rules.

    Hence {a}_{n } converges with a limit 0
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  2. #2
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    Quote Originally Posted by Arron View Post
    Is this correct?
    The dominant term is n!, so we write
    {a}_{n } = \frac{{n}^{4 }+2(n!) }{{n}^{3 }+{(-1)}^{n+1 }(n!) } = \frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } }
    Since { {n}^{4 }/n! },{ {n}^{3 }/n! } and { {(-1)}^{n+1 } } are all null sequences,
    \lim_{n \to \infty} {a}_{n } = \lim_{n \to \infty} \frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } } = \frac{0+2}{0+0 } = 0
    by the Combination Rules.
    Hence {a}_{n } converges with a limit 0
    Here is a problem with that: \lim _{n \to \infty } \left( { - 1} \right)^{n + 1}  \ne 0.
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  3. #3
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    Are you saying that { {(-1)}^{n+1 } } is not a basic null sequence. I think you are right. Does this mean this is not a converging sequence and how do I prove this?
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  4. #4
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    Quote Originally Posted by Arron View Post
    Are you saying that { {(-1)}^{n+1 } } is not a basic null sequence. I think you are right. Does this mean this is not a converging sequence and how do I prove this?
    Can you show that \left( {a_{2n} } \right) \to -2\;\& \,\left( {a_{2n - 1} } \right) \to   2~?
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  5. #5
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    No, thats a bit beyond me.

    I am thinking, | {(-1)}^{n+1 } | = 1

    Therefore the limit is \frac{0+2}{0+1 } = 2

    by the Combination Rules.
    Last edited by Arron; May 30th 2011 at 09:20 AM.
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  6. #6
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    Quote Originally Posted by Arron View Post
    No, thats a bit beyond me.
    Well you have a great deal of hard work ahead of you.
    Good Luck.
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