- #1

- 533

- 1

\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.

- Thread starter AxiomOfChoice
- Start date

- #1

- 533

- 1

\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.

- #2

- 430

- 3

[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

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- 0

- #4

- 430

- 3

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).

[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).

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