Can someone prove this series converges?

  • #1

Main Question or Discussion Point

[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.
 

Answers and Replies

  • #2
430
3
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.
 
  • #3
2
0
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.
 
  • #4
430
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.
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.
 
  • #5
1,631
4
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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