1. Apr 10, 2008

### motornoob101

For the factorial (2n+1)!, I thought the previous term is going to be (2(n-1)+1), which is equal to (2n-1).

Thus (2n+1)!= (2n+1)(2n-1)!

However, in the textbook, they have it as .

$$a_n= \frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n+1)(2n)(2n-1)!}$$

Are they wrong or I am wrong? Thanks!

2. Apr 10, 2008

### Diffy

The previous term of (2n+1) is (2n + 1) - 1 not (2(n-1) + 1).

In General if you have (f(x))! You can rewrite as f(x)*(f(x) - 1)!

What you tried which is incorrect is f(x)(f(x-1))!

See the difference?

3. Apr 10, 2008

### motornoob101

Ah ok. I see thanks. The reason I thought I was correct because I was looking at this example..

Which they are trying to determine if a series is convergent/divergent by the ratio test

Notice how they change 2n-1 to 2(n+1)-1? That's what confused me. Now I know they do it because it is the ratio test and you are trying to put $$a_{n+1}$$ but isn't that the same as what the factorial is doing? Thanks.

4. Apr 10, 2008

### Diffy

You are confusing terms in the sum, and terms within the factorial.