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Limit comparison test assistance needed please

  • Thread starter Randall
  • Start date
  • #1
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Homework Statement


Use the limit comparison test to check for convergence or divergence: Sum from n=1 to infinity of ((2n)^2+5)^-3

Homework Equations


let lim n to infinity of An/Bn = c
1) if 0<c<infinity then either both converge or both diverge
2) if c=0 and sum Bn converges, so does sum An
3) if c=infinity and sum Bn diverges, so does sum An

The Attempt at a Solution


see attached - I'm a little confused. It seems that it would be easy to pick a comparison series to "force" the answer that you want....
 

Attachments

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
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Chosing [itex] b_n = 1 + \frac{1}{\sqrt[3]{n} } [/itex] doesn't let you draw any conclusion because [itex] \sum_{n=1}^\infty b_n [/itex] doesn't converge.
 
  • #3
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Doesn't it converge to 1? 1 + 0 = 1? If not, why not? And also what should I choose instead to be bn? Thanks
 
  • #4
Stephen Tashi
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Doesn't it converge to 1? 1 + 0 = 1? If not, why not? And also what should I choose instead to be bn? Thanks
Don't confuse the convergence of the sequence [itex] \lim_{n \rightarrow \infty} b_n = 1 [/itex] with the convergence of the series [itex] \sum_{n=1}^\infty b_n [/itex]. The part 2 in your statement of the limit comparison tests talks about [itex] b_n [/itex] being summable.

My guess is [itex] b_n = (\frac{1}{n^2})^{-3} [/itex].

Edit: Try [itex] b_n = (n^2)^{-3} [/itex]
 
Last edited:
  • #5
epenguin
Homework Helper
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That looks like a photo taken with i-pad - for presentation the app mentioned below is almost no extra trouble and makes easier on the eye. :oldsmile: DocScan HD

They are the same ##L##. By the way, posting images instead of typing your work is frowned on in these forums. That goes double for images that are posted sideways.
"But if you do do that, there are apps, at least one I know called DocScan HD which if you are taking these with I-pad or similar cleans them up from the hard-to-read grey-yellow."
 
Last edited:
  • #6
epenguin
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If you find it very difficult to prove convergence, hedge your bets and try to test for divergence. I would compare with something like 1/n (which I hope you know is divergent.)
 
  • #7
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Thread closed. Please stop posting messy images of your work. Most of what you posted has been crossed out, and much of the rest is difficult to read.
 

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