ratio test (series) urgent test tommorow!

• December 2nd 2008, 02:50 PM
Legendsn3verdie
ratio test (series) urgent test tommorow!
i feel that the pic below explains my issue.

http://i176.photobucket.com/albums/w...titled-127.jpg
• December 2nd 2008, 02:57 PM
vincisonfire
n! means 1x2x3x...x(n-1)xn
Therefore
(n+1)! means 1x2x3x...x(n-1)xnx(n+1)
So $\frac{n!}{(n+1)!} = \frac{1x2x3x...x(n-1)xn}{x2x3x...x(n-1)xnx(n+1)} = \frac{1}{(n+1)}$
• December 2nd 2008, 03:17 PM
Legendsn3verdie
Quote:

Originally Posted by vincisonfire
n! means 1x2x3x...x(n-1)xn
Therefore
(n+1)! means 1x2x3x...x(n-1)xnx(n+1)
So $\frac{n!}{(n+1)!} = \frac{1x2x3x...x(n-1)xn}{x2x3x...x(n-1)xnx(n+1)} = \frac{1}{(n+1)}$

hm thanks kinda helps me but i m not fully understanding.. can any1 eleaborate?
• December 2nd 2008, 03:27 PM
vincisonfire
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
$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$
Here $\lim_{x \rightarrow \infty} \frac{n+2}{(n+1)^2} = \lim_{x \rightarrow \infty} \frac{1}{2(n+1)} =0$