Help verifying the series

**My Little Pony** · September 24th 2009, 10:20 PM

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?

**Prove It** · September 24th 2009, 10:23 PM

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 $\frac{x^n}{(2n - 1)!}$ .

**My Little Pony** · September 24th 2009, 10:24 PM

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!

**Prove It** · September 24th 2009, 10:27 PM

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

**My Little Pony** · September 24th 2009, 10:32 PM

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

**My Little Pony** · September 24th 2009, 11:04 PM

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?

**Prove It** · September 24th 2009, 11:41 PM

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

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

**My Little Pony** · September 24th 2009, 11:45 PM

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)!!

**mr fantastic** · September 25th 2009, 02:07 AM

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).

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