# Ratio test for infinite series

1. Feb 19, 2012

### bobsmith76

1. The problem statement, all variables and given/known data

Investigate the convergence of the following series.

(2n)!/(n!n!)

3. The attempt at a solution

Number one, I don't see how they get that
if an = (2n)!, then an+1 = ((2n+2))!, it should be (2n+1)!

Number two, I don't see how they go to the second step, why is the second step not:

(n!n!(2n+2)!)/((n+1)!(n+1)!(2n)!)? why do they have that extra (2n)! in the numerator?

Number three, I don't see why n!n! cancels in the numerator in step 3.

2. Feb 20, 2012

### SammyS

Staff Emeritus
Number one:
an+1 = ((2n+2))! results from basic substitution.

If $\displaystyle a_{n}= (2n)!\,,\ \text{ then }a_{n+1}=(2(n+1))!\,,\text{ so that }a_{n+1}=(2n+2)!$

Of course, this is not the same an as in the image you posted.

3. Feb 20, 2012

### SammyS

Staff Emeritus
$\displaystyle \frac{n!n!(2n+2)!}{(n+1)!(n+1)!(2n)!}=\frac{n!n!(2n+2)(2n+1)(2n)!}{(n+1)!(n+1)!(2n)!}$

This is because
$(2n+2)!=(2n+2)((2n+2)-1)((2n+2)-2)((2n+2)-3)((2n+2)-4)\dots(3)(2)(1)$
$=(2n+2)(2n+1)(2n)(2n-1)(2n-2)\dots(3)(2)(1)$

$=(2n+2)(2n+1)(2n)!$​

4. Feb 20, 2012

### SammyS

Staff Emeritus
$(n+1)!=(n+1)(n!)$

5. Feb 20, 2012

### bobsmith76

Great advice I really appreciate it.