Question

how do you determine the final terms in a telescoping series when the cancelling terms are not consecutive?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

You could post it, if you wanted ¬_¬

Answer 5

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

Answer 6

I understand how to do the partial fraction decomposition

Answer 7

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

Answer 8

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

Answer 9

\[s _{n}=1 + 1/2- (1/n-1) - (1/n)\]

Answer 10

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.

Answer 11

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

Answer 12

or the last two iterations

Answer 13

not parentheses

Answer 14

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

Answer 15

See cancelling in image

Answer 16

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

1/(n-1) - 1/(n+1)

Answer 17

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

Answer 18

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).

Answer 19

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.

Answer 20

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

Answer 21

No problem