## anonymous 5 years ago how do you determine the final terms in a telescoping series when the cancelling terms are not consecutive?

1. anonymous

Foolproof method: If you can't figure out a smart way, just write out the first 2/3/4 terms and check

2. anonymous

There will be better ways to figure it out, but that may be the most general/hardest to get wrong one.

3. anonymous

ok, I will try that now... stuck on a problem

4. anonymous

You could post it, if you wanted ¬_¬

5. anonymous

ok, that might help... $\sum_{n=2}^{\infty} 2/n ^{2}-1$

6. anonymous

I understand how to do the partial fraction decomposition

7. anonymous

I am getting stuck on the final terms when I sub in the series and cancel out

8. anonymous

how do I arrive to the following answer....

9. anonymous

$s _{n}=1 + 1/2- (1/n-1) - (1/n)$

10. anonymous

Write it out something like this (I would) 1/1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 ..... +1/(n-4) - 1/(n-2) +1/(n-3) - 1/(n-1) +1/(n-2) - 1/(n) If you start cancelling in pairs, you see that from the start only 1 + 1/2 don't cancel ... you could 'guess' that the second half of the last two will stay, but if you write out the last three too you can show this.

11. anonymous

so the last two terms in the last two parentheses stay?

12. anonymous

or the last two iterations

13. anonymous

not parentheses

14. anonymous

I don't see how you are coming up with the last 3 terms with the n...??

15. anonymous

See cancelling in image

16. anonymous

As for the last terms, just sub in n-2, n-1 and n to the partial fraction 1/(n-1) - 1/(n+1)

17. anonymous

I guess that is what I don't get... why am i subbing in n-2, n-1, and n? is that always?

18. anonymous

Well, if it's a sum from 2 to N, it will include 2, 3, 4 .... n-3, n-2, n-1 and n etc. Why do you ONLY go back to n-2? The difference in the partial fractions n-1 and n+1 is 2, so you can 'guess' that that will cancel back with the one two behind, so you need the last two at the end (but I used three so you can 'check' that they cancel how you think they will (see diagram).

19. anonymous

You could sub in the first 5 and the last 5 if you have enough time, just to make sure. Or none, if you can guess which will stay. Just depends on how confident you are.

20. anonymous

ok, I think I understand.... thank you, Mr. Newton :)

21. anonymous

No problem