i feel that the pic below explains my issue.

http://i176.photobucket.com/albums/w...titled-127.jpg

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- Dec 2nd 2008, 02:50 PMLegendsn3verdieratio test (series) urgent test tommorow!
i feel that the pic below explains my issue.

http://i176.photobucket.com/albums/w...titled-127.jpg - Dec 2nd 2008, 02:57 PMvincisonfire
n! means 1x2x3x...x(n-1)xn

Therefore

(n+1)! means 1x2x3x...x(n-1)xnx(n+1)

So $\displaystyle \frac{n!}{(n+1)!} = \frac{1x2x3x...x(n-1)xn}{x2x3x...x(n-1)xnx(n+1)} = \frac{1}{(n+1)} $ - Dec 2nd 2008, 03:17 PMLegendsn3verdie
- Dec 2nd 2008, 03:27 PMvincisonfire
Well the definition on n! is the product of integers up to n.

From this definition you can answer your question. I showed you that the product of integers up to n divided by the product of integers up to n+1 is equal to 1 over n+1.

Then to compute the limit, you have to use l'Hospital rule. That is

$\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} $

Here $\displaystyle \lim_{x \rightarrow \infty} \frac{n+2}{(n+1)^2} = \lim_{x \rightarrow \infty} \frac{1}{2(n+1)} =0 $