Did i do the reduction correctly? algebra type stuff!

  • Thread starter mr_coffee
  • Start date
  • #1
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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 [Broken]
Here is my work:
http://img203.imageshack.us/img203/9559/lastscan3jv.jpg [Broken]

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

  • #2
335
4
mr_coffee said:
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 [Broken]
Here is my work:
http://img203.imageshack.us/img203/9559/lastscan3jv.jpg [Broken]

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
 
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  • #3
789
4
I follow you until this line: [tex]\lim_{n\rightarrow\infty}\frac{(-x)(n^2+3)}{2n^2+4n+8}[/tex]

and then I don't see what you're doing.
 
  • #4
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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
 
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  • #5
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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, thats how i came out with -2 < x < 2
 
  • #6
789
4
I was wondering why you didn't just go straight to |x|/2. And yes, I think you're answer is correct.
 

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