Can someone prove this series converges?

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Discussion Overview

The discussion revolves around the convergence of the series \(\sum_{n=3}^\infty \frac{3}{n^2 - 4}\). Participants explore various methods to establish convergence, including the integral test, limit comparison test, and telescoping series.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that the integral test might work but finds it messy.
  • Another participant proposes testing the equivalent series \(\sum_{n=3}^\infty \frac{4}{n^2-4}\) and notes that it can be expressed as a telescoping series.
  • A participant mentions the limit comparison test with \(1/n^2\) as a potential method, indicating uncertainty about its ease.
  • Another participant agrees that the limit comparison test could be easier and acknowledges the utility of the telescoping argument for providing the actual limit.
  • One participant points out that \(\frac{3}{n^2-4}={\cal O}\left(\frac{1}{n^2}\right)\) for \(n>k\), suggesting the use of the comparison test.

Areas of Agreement / Disagreement

Participants express multiple competing views on the methods to prove convergence, with no consensus on a single approach being preferred.

Contextual Notes

Some methods discussed may depend on specific conditions or assumptions about the series, and participants do not resolve the effectiveness of each proposed method.

AxiomOfChoice
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[tex] \sum_{n=3}^\infty \frac{3}{n^2 - 4}.[/tex]

All the standard approaches I've used have failed. I think the integral test will work, but it's a bit messy.
 
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Well it's clearly equivalent to testing for convergence of:
[tex]\sum_{n=3}^\infty \frac{4}{n^2-4}[/tex]
so let's do this instead (the constants turn out nicer). We note:
[tex]\frac{4}{n^2-4} = \frac{1}{n-2}-\frac{1}{n+2}[/tex]
You can use this and a standard telescopping argument to show that it converges.
 
Not actually doing this, but couldn't you do the limit comparison test with 1/n^2? Maybe it's not as easy as it seems.
 
yuechen said:
Not actually doing this, but couldn't you do the limit comparison test with 1/n^2? Maybe it's not as easy as it seems.

Yes this would work just as well. I used my approach mainly because it also gives you the actual limit (whether you care for it or not), and it uses more elementary tools.

I guess the limit comparison test is actually a bit easier to carry out now that you point it out and it may have been the intended solution.
 
As already pointed out, you could use the comparison test, and the fact that :


[tex]\frac{3}{n^2-4}={\cal O}\left(\frac{1}{n^2}\right)[/tex],

for n>k (in this case k=4 would work).
 
Last edited:

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