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

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

Mathematics
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Foolproof method: If you can't figure out a smart way, just write out the first 2/3/4 terms and check
There will be better ways to figure it out, but that may be the most general/hardest to get wrong one.
ok, I will try that now... stuck on a problem

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You could post it, if you wanted ¬_¬
ok, that might help... \[\sum_{n=2}^{\infty} 2/n ^{2}-1\]
I understand how to do the partial fraction decomposition
I am getting stuck on the final terms when I sub in the series and cancel out
how do I arrive to the following answer....
\[s _{n}=1 + 1/2- (1/n-1) - (1/n)\]
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.
so the last two terms in the last two parentheses stay?
or the last two iterations
not parentheses
I don't see how you are coming up with the last 3 terms with the n...??
See cancelling in image
1 Attachment
As for the last terms, just sub in n-2, n-1 and n to the partial fraction 1/(n-1) - 1/(n+1)
I guess that is what I don't get... why am i subbing in n-2, n-1, and n? is that always?
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).
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.
ok, I think I understand.... thank you, Mr. Newton :)
No problem

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