# Thread: Series Convergence Question i.e. Ratio Test

1. ## Series Convergence Question i.e. Ratio Test

I basically need help with the algebra involved in this problem. Specifically an explanation of how the ! and the 2^n are going to cancel out. Thank You.

2. Originally Posted by ihatemath
I basically need help with the algebra involved in this problem. Specifically an explanation of how the ! and the 2^n are going to cancel out. Thank You.

ratio test ...

$\displaystyle \displaystyle \lim_{n \to \infty} \frac{(n+2)!}{2^{n+1}} \cdot \frac{2^n}{(n+1)!}$

note that $\displaystyle \displaystyle \frac{(n+2)!}{(n+1)!} = n+2$

and the limit simplifies to ...

$\displaystyle \displaystyle \lim_{n \to \infty} \frac{n+2}{2}$

3. Yes, I got that far and got stuck as to why the second step you posted = n+2. I don't understand how you got from point A to point B

4. Originally Posted by ihatemath
Yes, I got that far and got stuck as to why the second step you posted = n+2.
Because $\displaystyle n! = n(n-1)! = n(n-1)(n-2)! = n(n-1)(n-2)(n-3)!$ etc.
i.e. $\displaystyle n! = \prod_{k=0}^{n-1}(n-k).$ So $\displaystyle (n+2)! = (n+2)(n+2-1) = (n+2)(n+1)!$

5. $\displaystyle \displaystyle \frac{(n+2)!}{(n+1)!} = \frac{(n+2)(n+1)(n)(n-1)(n-2) ... (3)(2)(1)}{(n+1)(n)(n-1)(n-2) ... (3)(2)(1)}$