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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Foolproof method: If you can't figure out a smart way, just write out the first 2/3/4 terms and check

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There will be better ways to figure it out, but that may be the most general/hardest to get wrong one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, I will try that now... stuck on a problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could post it, if you wanted ¬_¬

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, that might help... \[\sum_{n=2}^{\infty} 2/n ^{2}1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I understand how to do the partial fraction decomposition

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am getting stuck on the final terms when I sub in the series and cancel out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do I arrive to the following answer....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[s _{n}=1 + 1/2 (1/n1)  (1/n)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Write it out something like this (I would) 1/1  1/3 + 1/2  1/4 + 1/3  1/5 ..... +1/(n4)  1/(n2) +1/(n3)  1/(n1) +1/(n2)  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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the last two terms in the last two parentheses stay?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or the last two iterations

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't see how you are coming up with the last 3 terms with the n...??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0See cancelling in image

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0As for the last terms, just sub in n2, n1 and n to the partial fraction 1/(n1)  1/(n+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess that is what I don't get... why am i subbing in n2, n1, and n? is that always?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, if it's a sum from 2 to N, it will include 2, 3, 4 .... n3, n2, n1 and n etc. Why do you ONLY go back to n2? The difference in the partial fractions n1 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, I think I understand.... thank you, Mr. Newton :)
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