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

In summary, the series 2*4*6*8*(2n)/2*4*6*8...(2n+2) can be simplified using the formula 2^n*n! and can be reduced to 1/(2n+2) by factoring out the common factor of 2 from each term in the series.
  • #1
wuffle
26
0

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 :(
 
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  • #2
wuffle said:

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! ?" .

[itex]2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\dots(2n-2)\cdot(2n)[/itex]
[itex]=(2\cdot1)(2\cdot2)(2\cdot3)(2\cdot4)(2\cdot5)(2 \cdot6)(2\cdot7)\dots2(n-1)\cdot2(n)[/itex]

[itex]=2^n\cdot1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7 \dots(n-1)\cdot(n)[/itex]

...​
 
  • #3
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.
 
  • #4
wuffle said:

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.
 
  • #5
thanks for replies! appreciate it
 

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

What is the pattern in this series?

The pattern in this series is that each term is the product of the previous term and an even number. The even number increases by 2 with each term.

What is the general formula for this series?

The general formula for this series is 2*4*6*8*...(2n), where n is the number of terms in the series.

What is the simplified form of this series?

The simplified form of this series is 2^n, where n is the number of terms in the series.

How do you find the sum of this series?

To find the sum of this series, we can use the formula for the sum of a geometric series: S = a(1-r^n) / (1-r), where a is the first term, r is the common ratio, and n is the number of terms. In this case, a = 2, r = 2, and n is the number of terms. So the sum of this series is 2*(1-2^n) / (1-2) = 2*(1-2^n) / (-1) = 2^(n+1) - 2.

What is the purpose of simplifying this series?

The purpose of simplifying this series is to make it easier to understand and work with. The simplified form allows us to quickly find the sum of the series and make predictions about future terms without having to write out and calculate each individual term. It also allows us to see the pattern and make generalizations about the series.

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