Can someone prove this series converges?

[AxiomOfChoice]

<br /> \sum_{n=3}^\infty \frac{3}{n^2 - 4}.<br />

All the standard approaches I've used have failed. I think the integral test will work, but it's a bit messy.

 

[rasmhop]

Well it's clearly equivalent to testing for convergence of:
\sum_{n=3}^\infty \frac{4}{n^2-4}
so let's do this instead (the constants turn out nicer). We note:
\frac{4}{n^2-4} = \frac{1}{n-2}-\frac{1}{n+2}
You can use this and a standard telescopping argument to show that it converges.

 

[yuechen]

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.

 

[rasmhop]

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.

 

[sutupidmath]

As already pointed out, you could use the comparison test, and the fact that :


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

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