Simplifying Series: 2*4*6*8*(2n)/2*4*6*8...(2n+2)

[wuffle]

Homework Statement

This is a sub problem, the whole problem is find interval of convergence and radius of convergence but the thing I am having trouble with is how to simplify this
simplify the series

2*4*6*8*(2n)/2*4*6*8...(2n+2)

Homework Equations

The Attempt at a Solution

keys say 2*4*6*8...(2n)=2^n*n!

so 2^n*n!/2^(n+1)*(n+1)!=1/(2n+2)

my question is, HOW? how do u go from 2*4*6*8...(2n) to 2^n*n! or how do u get from

2*4*6*8*(2n)/2*4*6*8...(2n+2) to 1/(2n+2)

sorry for not using the proper tools, they are very confusing :(

 

[SammyS]

Homework Statement

This is a sub problem, the whole problem is find interval of convergence and radius of convergence but the thing I am having trouble with is how to simplify this
simplify the series

2*4*6*8*(2n)/2*4*6*8...(2n+2)

Homework Equations

The Attempt at a Solution

keys say 2*4*6*8...(2n)=2^n*n!

so 2^n*n!/2^(n+1)*(n+1)!=1/(2n+2)

my question is, HOW? how do u go from 2*4*6*8...(2n) to 2^n*n! or how do u get from

2*4*6*8*(2n)/2*4*6*8...(2n+2) to 1/(2n+2)

sorry for not using the proper tools, they are very confusing :(

I'll address, "HOW? how do u go from 2*4*6*8...(2n) to 2^n*n! ?" .

$2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\dots(2n-2)\cdot(2n)$

$=(2\cdot1)(2\cdot2)(2\cdot3)(2\cdot4)(2\cdot5)(2 \cdot6)(2\cdot7)\dots2(n-1)\cdot2(n)$

$=2^n\cdot1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7 \dots(n-1)\cdot(n)$

...​

 

[Dick]

For example, 2*4*6*8=(2*1)*(2*2)*(2*3)*(2*4)=(2*2*2*2)*(1*2*3*4)=2^4*4!. Just think about it a little. It's not magic.

 

[LCKurtz]

Homework Statement

This is a sub problem, the whole problem is find interval of convergence and radius of convergence but the thing I am having trouble with is how to simplify this
simplify the series

2*4*6*8*(2n)/2*4*6*8...(2n+2)

Homework Equations

The Attempt at a Solution

keys say 2*4*6*8...(2n)=2^n*n!

so 2^n*n!/2^(n+1)*(n+1)!=1/(2n+2)

my question is, HOW? how do u go from 2*4*6*8...(2n) to 2^n*n!

2*4*6*8...*(2n) = (1*2)*(2*2)*(3*2)*(4*2)...(n*2) The first factor in each parentheses gives the n! and the second gives 2ⁿ.

or how do u get from

2*4*6*8*(2n)/2*4*6*8...(2n+2) to 1/(2n+2)

sorry for not using the proper tools, they are very confusing :(

In the denominator the factors increase by 2 each factor. What would the factor right in front of the (2n+2) be?

[Edit]Guess I have to learn to type faster. While I'm typing, two other responses appear.

 

[wuffle]

thanks for replies! appreciate it