1. ## Trouble proving convergence/divergence

I've got a series which i applied the ratio test to, after simplifying a bit im left with:

(r+1)*5*(2r)!
(2(r+1))!

What should be my next step? i was thing of dividing by (2(r+1))! before tending r to inf.
But that leaves you with exact same equation, leaving it like that it would appear the equation tends to 0.
(sorry for lack of correct math symbols)

Thanks

2. Originally Posted by sj247
I've got a series which i applied the ratio test to, after simplifying a bit im left with:

(r+1)*5*(2r)!
(2(r+1))!

What should be my next step? i was thing of dividing by (2(r+1))! before tending r to inf.
But that leaves you with exact same equation, leaving it like that it would appear the equation tends to 0.
(sorry for lack of correct math symbols)

Thanks

$\displaystyle \frac{(2r)!}{(2r+2)!}=\frac{1}{(2r+1)(2r+2)}$

$\displaystyle \dfrac{5(r+1)(2r)!}{(2(r+1))!}=\dfrac{5(r+1)(2r)!} {(2r+2)!}=\dfrac{5(r+1)(2r)!}{(2r+2)(2r+1)(2r)!}\d ots$

4. Originally Posted by sj247
I've got a series which i applied the ratio test to, after simplifying a bit im left with:

(r+1)*5*(2r)!
(2(r+1))!

What should be my next step? i was thing of dividing by (2(r+1))! before tending r to inf.
But that leaves you with exact same equation, leaving it like that it would appear the equation tends to 0.
(sorry for lack of correct math symbols)

Thanks
Note that (2(r+1))! = (2r + 2)! = (2r+2)(2r+1)(2r)! so yes, the limit is zero and a conclusion can be made ....

5. Thanks guys . I got similar in my working, just needed to check.

6. Originally Posted by Ackbeet
$\displaystyle \dfrac{5(r+1)(2r)!}{(2(r+1))!}=\dfrac{5(r+1)(2r)!} {(2r+2)!}=\dfrac{5(r+1)(2r)!}{(2r+2)(2r+1)(2r)!}\d ots$
$\displaystyle \displaystyle = \frac{5(r+1)(2r)!}{2(r+1)(2r+1)(2r)!}$...