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Can someone prove this series converges?

  1. Apr 1, 2010 #1
    \sum_{n=3}^\infty \frac{3}{n^2 - 4}.

    All the standard approaches I've used have failed. I think the integral test will work, but it's a bit messy.
  2. jcsd
  3. Apr 1, 2010 #2
    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.
  4. Apr 1, 2010 #3
    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.
  5. Apr 1, 2010 #4
    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.
  6. Apr 2, 2010 #5
    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: Apr 2, 2010
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