Did i do the reduction correctly? algebra type stuff

[mr_coffee]

Directions:
Find all the values of x such that the given series would converge.

http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/fd/74dc1166ac21d3f59ee568a71eb55a1.png
Here is my work:
http://img203.imageshack.us/img203/9559/lastscan3jv.jpg

I got it wrong but is it because i don't know how to do inequalities, or just suck in general?

 

[topsquark]

Directions:
Find all the values of x such that the given series would converge.

http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/fd/74dc1166ac21d3f59ee568a71eb55a1.png
Here is my work:
http://img203.imageshack.us/img203/9559/lastscan3jv.jpg

I got it wrong but is it because i don't know how to do inequalities, or just suck in general?

What happened to the 2?

-Dan

 

[Jameson]

I follow you until this line: $$\lim_{n\rightarrow\infty}\frac{(-x)(n^2+3)}{2n^2+4n+8}$$

and then I don't see what you're doing.

 

[mr_coffee]

Okay from here, I'm using the rule, where u divide by the largest exponent, in this case its n^2, so u divide every term by n^2 and leave the -x alone, then u take as n->infinity. I'm using this website as my guide:
http://www.math.unh.edu/~jjp/radius/radius.html

Ooo i don't know where the 2 went lol i will do it again with the 2

 

[mr_coffee]

Sweet it worked once i added the 2 lol.

|-x|/2 < 1

|-x| < 2
-2 < x <2

quick question though. is |-x| == |x|? i just assumed it was, that's how i came out with -2 < x < 2

 

[Jameson]

I was wondering why you didn't just go straight to |x|/2. And yes, I think you're answer is correct.