Thread: ratio test (series) urgent test tommorow!

1. ratio test (series) urgent test tommorow!

i feel that the pic below explains my issue.

2. n! means 1x2x3x...x(n-1)xn
Therefore
(n+1)! means 1x2x3x...x(n-1)xnx(n+1)
So $\displaystyle \frac{n!}{(n+1)!} = \frac{1x2x3x...x(n-1)xn}{x2x3x...x(n-1)xnx(n+1)} = \frac{1}{(n+1)}$

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

hm thanks kinda helps me but i m not fully understanding.. can any1 eleaborate?

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