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## Homework Statement

I want to simplify the following (these are to do with the ratio test in series):

a) [tex]\frac{(n+1)!}{(2n+1)}/\frac{n!}{2n-1}[/tex]

b) [tex]\frac{2^{2n+1}}{(2n+1)!} / \frac{2^{2n-1}}{(2n-1)!}[/tex]

## The Attempt at a Solution

a) [tex]\frac{(n+1)!}{(2n+1)}/\frac{n!}{2n-1}[/tex]

[tex]\frac{(n+1)!}{(2n+1)} \times \frac{(2n-1)}{n!}[/tex]

So, we devide up the factorial terms: [tex]\frac{n \times n+1}{n} = n+1[/tex]

and deviding the rest:

[tex]\frac{2n-1}{2n+1}=-1[/tex]

Therefore the whole thing should simplify to: -(n+1) But the book says it must be [tex]\frac{n+1}{2n+1}[/tex], what's wrong?

b) [tex]\frac{2^{2n+1}}{(2n+1)!} / \frac{2^{2n-1}}{(2n-1)!}[/tex]

[tex]\frac{2^{2n+1}}{(2n+1)!} \times \frac{2^{2n-1}}{(2n-1)!}[/tex]

[tex]\frac{4n . 2}{4n . 1/2} = 4[/tex]

(2n-1)!/(2n+1)! = [tex]\frac{(2n)(2n-1)}{(2n)(2n+1)}[/tex]

So based on what I've done my answer has to be -4 but the real answer has to be:

4/(2n+1)(2n)

Can anyone please help me with this simplifications...