Results 1 to 2 of 2

Math Help - 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.

    <br />
a_n  = \frac{{(2n + 3)!}}<br />
{{(n + 1)!}}<br />

    So...

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

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

    <br />
(2n + 5)(2n + 4) > n + 2<br />

    It also says that is not bounded because:

    <br />
\frac{{(2n + 3)!}}<br />
{{(n + 1)!}} = (2n + 3)(2n + 2)...(n + 2)<br />

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

    Thank you.
    Follow Math Help Forum on Facebook and Google+

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

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

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

    <br />
\frac{(n+2)!}{{(n + 1)!}} < \frac{{(2n + 5)!}}<br />
{{(2n + 3)!}} \hfill \\ <br />

    LHS :

    \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 :

    \frac{{(2n + 5)!}}<br />
{{(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:

    <br />
\frac{{(2n + 3)!}}<br />
{{(n + 1)!}} = (2n + 3)(2n + 2)...(n + 2)<br />
    Same idea as before : simplify.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 24th 2010, 02:10 AM
  2. sequence problem
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 17th 2009, 11:57 AM
  3. Sequence Problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 27th 2008, 07:48 PM
  4. Sequence Problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 19th 2008, 11:29 AM
  5. A sequence problem.
    Posted in the Calculus Forum
    Replies: 9
    Last Post: January 13th 2008, 04:25 PM

Search Tags


/mathhelpforum @mathhelpforum