Ratio test for infinite series

[bobsmith76]

Homework Statement

Investigate the convergence of the following series.

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

The Attempt at a Solution

Number one, I don't see how they get that
if a_n = (2n)!, then a_n+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.

 

[SammyS]

Homework Statement

Investigate the convergence of the following series.

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

The Attempt at a Solution

Number one, I don't see how they get that
if a_n = (2n)!, then a_n+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.

Number one:
a_n+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 a_n as in the image you posted.

 

[SammyS]

...

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?

...

\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)!​

 

[SammyS]

...

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

(n+1)!=(n+1)(n!)

 

[bobsmith76]

Great advice I really appreciate it.