Why Does αn+1 Include an Extra Factor in the Denominator?

[shanepitts]

Below is a screen shot of a solution to a problem. The part I don't fathom is after the ratio test is applied to the denominator. How can, noting that a_n+1, (2n-1) become (2n-1)(2n+1) and not just (2(n+1)-1)=2n+1?

Thank you in advance

 

[Mark44]

Below is a screen shot of a solution to a problem. The part I don't fathom is after the ratio test is applied to the denominator. How can, noting that a_n+1, (2n-1) become (2n-1)(2n+1) and not just (2(n+1)-1)=2n+1?

Thank you in advance
View attachment 83898

In a_n, the denominator is the product $1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)$. What will be the next factor in the denominator for a_{n + 1}?

BTW, when you post a question here, please don't delete the three parts of the homework template.

 

[shanepitts]

In a_n, the denominator is the product $1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)$. What will be the next factor in the denominator for a_{n + 1}?

BTW, when you post a question here, please don't delete the three parts of the homework template.

Thanks for the quick reply

Asked in that manner, the next part is 2n+1. But why not just replace the n with (n+1)? And why would the answer be completely different if this approach to problem is taken?

 

[Mark44]

Thanks for the quick reply

Asked in that manner, the next part is 2n+1. But why not just replace the n with (n+1)? And why would the answer be completely different if this approach to problem is taken?

It wouldn't. If you replace n by n + 1 in the expression 2n - 1, what do you get?

 

[shanepitts]

It wouldn't. If you replace n by n + 1 in the expression 2n - 1, what do you get?

You get 2n+1 instead of the (2n-1)(2n+1), no?

 

[Mark44]

You get 2n+1 instead of the (2n-1)(2n+1), no?

You get (2n + 1). The (2n - 1) factor is the one from a_n.

 

[shanepitts]

You get (2n + 1). The (2n - 1) factor is the one from a_n.

Thanks,
I'm sorry but I still don't fathom.

Let's say

α_n=2n-1 then α_n+1 should equal 2(n+1)-1=2n+1

Why the extra 2n-1 in the α_n+1?

 

[Mark44]

Thanks,
I'm sorry but I still don't fathom.

Let's say

α_n=2n-1 then α_n+1 should equal 2(n+1)-1=2n+1

This is not a_n, at least as it's defined in post 1.

$$a_n = \frac{x^n}{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}$$

Now, what is a_{n + 1}?
You need to ask yourself how many factors are in the denominator of a_n? How many are in the denominator of a_{n + 1}?

Why the extra 2n-1 in the α_n+1?