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Thread: Sequence Problem

  1. #1
    Member RedBarchetta's Avatar
    Joined
    Apr 2008
    From
    United States
    Posts
    114

    Sequence Problem

    Determine if the sequence is nondecreasing and if it's bounded from above.

    $\displaystyle
    a_n = \frac{{(2n + 3)!}}
    {{(n + 1)!}}
    $

    So...

    $\displaystyle
    \begin{gathered}
    a_n \leqslant a_{n + 1} \hfill \\
    \frac{{(2n + 3)!}}
    {{(n + 1)!}} < \frac{{(2(n + 1) + 3)!}}
    {{((n + 1) + 1)!}} \hfill \\
    \frac{{(2n + 3)!}}
    {{(n + 1)!}} < \frac{{(2n + 5)!}}
    {{(n + 2)!}} \hfill \\
    \frac{{(n + 2)!}}
    {{(n + 1)!}} < \frac{{(2n + 5)!}}
    {{(2n + 3)!}} \hfill \\
    \end{gathered}
    $

    Then I have no clue how the answer key got to this:

    $\displaystyle
    (2n + 5)(2n + 4) > n + 2
    $

    It also says that is not bounded because:

    $\displaystyle
    \frac{{(2n + 3)!}}
    {{(n + 1)!}} = (2n + 3)(2n + 2)...(n + 2)
    $

    I don't quite follow what's going on there.

    Thank you.
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  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,
    Quote Originally Posted by RedBarchetta View Post
    (...)
    $\displaystyle
    \frac{(n+2)!}{{(n + 1)!}} < \frac{{(2n + 5)!}}
    {{(2n + 3)!}} \hfill \\
    $

    Then I have no clue how the answer key got to this:

    $\displaystyle
    (2n + 5)(2n + 4) > n + 2
    $
    Remember the definition of $\displaystyle n!$ : $\displaystyle n! = n\times(n-1)\times(n-2)\cdots 3\times 2\times 1$. Using this, one can simplify both sides of

    $\displaystyle
    \frac{(n+2)!}{{(n + 1)!}} < \frac{{(2n + 5)!}}
    {{(2n + 3)!}} \hfill \\
    $

    LHS :

    $\displaystyle \frac{(n+2)!}{{(n + 1)!}}=\frac{(n+2)\times{\color{blue}(n+1)\times n \times (n-1)\cdots \times 2 \times 1}}{{\color{blue}(n+1)\times n \times (n-1)\cdots \times 2 \times 1}}=n+2$

    RHS :

    $\displaystyle \frac{{(2n + 5)!}}
    {{(2n + 3)!}} = \frac{(2n+5)\times(2n+4)\times{\color{blue}(2n+3)\ times(2n+2)\cdots2\times 1}}{{\color{blue}(2n+3)\times(2n+2)\cdots 2\times 1}}=(2n+5)(2n+4)$

    It also says that is not bounded because:

    $\displaystyle
    \frac{{(2n + 3)!}}
    {{(n + 1)!}} = (2n + 3)(2n + 2)...(n + 2)
    $
    Same idea as before : simplify.
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