Limit comparison test assistance needed please

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Homework Help Overview

The original poster attempts to apply the limit comparison test to determine the convergence or divergence of the series given by the sum from n=1 to infinity of ((2n)^2+5)^-3. The discussion revolves around selecting an appropriate comparison series.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various choices for the comparison series, with some questioning the validity of their selections. There is confusion regarding the convergence of the sequence versus the convergence of the series.

Discussion Status

Participants are exploring different comparison series and questioning the assumptions behind their choices. Some guidance has been offered regarding the selection of comparison series, but no consensus has been reached on the best approach.

Contextual Notes

There are indications of confusion regarding the convergence of sequences versus series, as well as concerns about the presentation of the original poster's work. The discussion also highlights the challenges of proving convergence and the potential to test for divergence instead.

Randall
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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

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

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

Edit: Try b_n = (n^2)^{-3}
 
Last edited:
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

LCKurtz said:
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:
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.)
 
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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