# Can someone prove this series converges?

$$\sum_{n=3}^\infty \frac{3}{n^2 - 4}.$$

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

## Answers and Replies

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.

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.

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 :

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

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

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