# Sequence Problem

• Oct 11th 2008, 03:41 PM
RedBarchetta
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.
• Oct 11th 2008, 11:59 PM
flyingsquirrel
Hi,
Quote:

Originally Posted by RedBarchetta
(...)
$\displaystyle \frac{(n+2)!}{{(n + 1)!}} < \frac{{(2n + 5)!}} {{(2n + 3)!}} \hfill \\$

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 :

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

Quote:

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.