# Thread: Help verifying the series

1. ## Help verifying the series

Hi, I'm not sure how to interpret this series so I was wondering if someone could clarify what it means.

x^n/(1*3*5*...*(2n-1))

Is the denominator just (2n-1)? Or do I separate the (2n-1) from the other numbers? If the latter, it would become (2n-1)^2 correct?

2. Originally Posted by My Little Pony
Hi, I'm not sure how to interpret this series so I was wondering if someone could clarify what it means.

x^n/(1*3*5*...*(2n-1))

Is the denominator just (2n-1)? Or do I separate the (2n-1) from the other numbers? If the latter, it would become (2n-1)^2 correct?
Note that $1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = (2n - 1)!$.

So the series is $\frac{x^n}{(2n - 1)!}$.

3. Originally Posted by Prove It
Note that $1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = (2n - 1)!$.

So the series is $\frac{x^n}{(2n - 1)!}$.
Ah gotcha! Thanks a bunch!

4. Do you need to work out if it converges? If so, the ratio test should do nicely...

5. I'm supposed to find the radius and interval of convergence. But it should be much easier now that I have that cleared up.

6. Originally Posted by Prove It
Note that $1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = (2n - 1)!$.

So the series is $\frac{x^n}{(2n - 1)!}$.
Are you sure this is correct?

I thought (2n-1)! = (2n-1)(2n-2)(2n-3)...3x2x1?

7. Originally Posted by My Little Pony
Are you sure this is correct?

I thought (2n-1)! = (2n-1)(2n-2)(2n-3)...3x2x1?
Any odd number can be written as $2n - 1$ if $n \in \mathbf{N}$.

If $n = 1$ then $2n - 1 = 1$.

If $n = 2$ then $2n - 1 = 3$.

If $n = 3$ then $2n - 1 = 5$. Etc.

The counter is not $2n - 1$. The counter is $n$.

So, letting $n \in \mathbf{N}$ we have

$(2n - 1)! = 1\times 3 \times 5 \times 7 \times \dots \times (2n - 5)\times(2n - 3)\times (2n - 1)$.

8. Originally Posted by Prove It
Any odd number can be written as $2n - 1$ if $n \in \mathbf{N}$.

If $n = 1$ then $2n - 1 = 1$.

If $n = 2$ then $2n - 1 = 3$.

If $n = 3$ then $2n - 1 = 5$. Etc.

The counter is not $2n - 1$. The counter is $n$.

So, letting $n \in \mathbf{N}$ we have

$(2n - 1)! = 1\times 3 \times 5 \times 7 \times \dots \times (2n - 5)\times(2n - 3)\times (2n - 1)$.
My tutor informs me that what you're talking about is the double factorial of (2n-1).

(2n-1)!!

9. Originally Posted by Prove It
Note that $1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = (2n - 1)!$.

[snip]
No. In fact:

$1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = \frac{(2n)!}{n! 2^n}$

(easily proved by induction).