series--n! again

C

chopchop

New member
Joined
May 4, 2006
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11
Thanks to evereyone for their thorough help, i love u guys. I have one more question-last one i promise.

this problem:

1+2x+(4!x^2)/(2!)^2 + (6!x^3)/(3!)^2+(8!x^4)/(4!)^2+(10!x^5)/(5!)^2+...

i know that using the ratio test:


[ 2(n+1)! / (n+1)!^2 ] X^(n+1) ]
_______________________________

[ (2n)! / (n)! ^2] X^n


The part with the Xs I get, what i don't get though
how to arrive at this answer

[ 2(n+1)! / (n+1)!^2 ]
______________________..........=(2n+2)(2n+1)/(n+1)^2

[ (2n)! / (n)! ^2]


I keep getting confused with these Ns....It's probably very stupid to ask, but can someone show me step by step?
Thank u so much
 
I do not understand

[ 2(n+1)! / (n+1)!^2 ]
______________________..........=(2n+2)(2n+1)/(n+1)^2

[ (2n)! / (n)! ^2]

i'm sorry, i know it's pretty stupd i said....but
a. i get 2(n+1)=2n+2
b. i get that (n+1)^2 stays
c. but what happened with 2n! and n^2....i just don't get how u get the right side, enlighten me....i'm still confused on this part...until pretty recently have never seen n! in my life.
 
For clarity of formatting, please consider learning a little LaTeX. Using simple coding like "\frac{n!}{10^n}" can make your text look like \(\displaystyle \frac{n!}{10^n}\)
__________
Edit: Links for learning LaTeX formatting are available in the "Forum Help" pull-down menu at the very top of the page. -- stapel
 
Look you have to do something for yourself!
\(\displaystyle \L
(2n + 2)! = (2n + 2)(2n + 1)(2n)!\)
 
As I stated before, try looking at your factorial like this. You know what a factorial is, right?.

So, if you have \(\displaystyle (2n+2)!\), then that equals

\(\displaystyle \L\\(2n+2)(2n+1)(2n)(2n-1)(2n-2)(2n-3)..........\)

\(\displaystyle \L\\((n+1)!)^{2}=(n+1)^{2}n^{2}(n-1)^{2}(n-2)^{2}(n-3)^{2}.....\)

\(\displaystyle \L\\(n!)^{2}=n^{2}(n-1)^{2}(n-2)^{2}(n-3)^{2}................\)

\(\displaystyle \L\\((2n)!)=2n(2n-1)(2n-2)(2n-3)..................\)

So, you have:

\(\displaystyle \L\\\(\frac{(2n+2)(2n+1)\sout{(2n)}\sout{(2n-1)}\sout{(2n-2)}\sout{(2n-3)}..........}{(n+1)^{2}\sout{n^{2}(n-1)^{2}(n-2)^{2}(n-3)^{2}....}})\cdot\(\frac{\sout{n^{2}(n-1)^{2}(n-2)^{2}(n-3)^{2}}................}{\sout{2n(2n-1)(2n-2)(2n-3)}.................})\)

Now, cancel what cancels. As you can hopefully see, you are left with what you want. Sorry about the parentheses. I'm having a rough time getting LaTex to display them correctly.
 
chopchop said:
until pretty recently have never seen n! in my life.
Click to expand...

If indeed you have not ever seen factorials before and you are in a calculus class doing series then you have been horribly advised. Go to your advisor or academic dean and complain in the strongest terms. You are misplaced. You do not have the background for this work.

No one here, certainly not I, has anything but hope for your learning mathematics.
 
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