does anyone know how 2n+2! becomes (2n+2)(2n+1)

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- anonymous

does anyone know how 2n+2! becomes (2n+2)(2n+1)

- jamiebookeater

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- anonymous

i guess it must b (2n +2)! ?

- anonymous

yah

- anonymous

n! = n(n-1)!

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## More answers

- anonymous

(2n+2)(2n+1) (2n)....1

- anonymous

(2n+2)!= (2n+2)(2n+1)!

- anonymous

on cramster one of the problem shows it changes from (2n+2)! and becomes (2n+1)(2n+2) with no factorial sign behind it

- anonymous

there is a facatorial behind it......

- anonymous

i mean they don't show it like this (2n+2)(2n+1) (2n)....1

- anonymous

I meant that u keep decreasing by 1 until u reach 1
another way of writing it (2n+2)(2n+1) (2n)...3.2.1

- anonymous

yah they dont show it decreasing it stops after 2n+1

- anonymous

yes but there must be a factorial with it

- anonymous

and there isnt

- anonymous

As far as I know about factorials ...there should be...I could be wrong

- anonymous

is there anyway you can look at the problem on cramster

- anonymous

u can check on wolframalpha.com

- anonymous

there must be factorial with it

- anonymous

there has to be a factorial behind it, unless there is something else about your problem... cramster sometimes has mistakes

- anonymous

the things is my teacher did the same thing so im confused

- anonymous

the only way the factorial can drop out is if it cancelled out in a fraction

- anonymous

then check did u post the right question?

- anonymous

this is the actual problem determine of this series converges or diverges\[\sum_{n=1}^{\infty} (n!)^2/(2n) !\]

- anonymous

use ratio test

- anonymous

n u applied some test for the convergence of series?

- anonymous

right

- anonymous

i did but i didnt correct answer if someone can please show me how to do this it can help me with the rest of my problems

- anonymous

sigh i hate it when people post half the question... thigns cancel and become (n+1)^2/(2n+2)(2n+1)

- anonymous

what is 2n!

- anonymous

yes it converges, so you use ratio test...
an+1/an= [ ((n+1)!)^2/(2(n+1))! ]/ [ (n!)^2/(2n)! ]
flipping fractions to get 1 whole one..
(2n)!(n+1)!(n+1)!/(n!)(n!)(2n+2)!
which is where your question arises, like I said, the only reason the factorial disappeared is cause it got cancelled, which you always want to do when doing ratio test, so you simply keep pulling your factorials apart until they can cancel
so on top (n+1)!=(n+1)(n!), which you do twice to get rid of both the n! on the bottom
then on the bottom you have (2n+2)! so you pull the factorial apart twice to get (2n+2)(2n+1)(2n)! then the (2n)! can cancel with the one on top and you get left with (n+1)^2/(2n+2)(2n+1). take the limit and it should be 1/4 so it converges

- anonymous

yes i was able to catch my mistake thank you very much. i wanted to know how does 2n! decrease

- anonymous

does it become (2n)(2n-1)(2n-2)..1

- anonymous

yes

- anonymous

hey thank u for ur help im gonna become a fan

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