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precal
 4 years ago
sequence problem _ Calculus
precal
 4 years ago
sequence problem _ Calculus

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precal
 4 years ago
Best ResponseYou've already chosen the best response.0don't I need to use some formula?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0So is the question, "Does the series converge?" ? :o

precal
 4 years ago
Best ResponseYou've already chosen the best response.0yes, I know it converges to 9/4

precal
 4 years ago
Best ResponseYou've already chosen the best response.0doesn't help that I don't know why?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351477186152:dw Hmm can't you do some type of comparison test? Sorry I'm a little rusty with these :D We know that the pseries 1/n^2 converses, and our series is smaller than that one :O something like that..

precal
 4 years ago
Best ResponseYou've already chosen the best response.0how about if I do this as a telescoping series? as if I know what I am talking about

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0Yah that's prolly the way to approach it c: try to recognize a pattern coming out of it.

precal
 4 years ago
Best ResponseYou've already chosen the best response.0I know if I factor out the 9 and then I just need to create the (1/4)

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0Oh oh oh i think it's coming back to me a little bit... I think in order to write it as a telescoping series, we need to rewrite it as partial fractions, then we'll get terms being subtracted in each n, allowing for some cancellations probably.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351479486392:dw

precal
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, I ran off to watch some youtube videos on this

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0Mmmm that's prolly a good idea :) I don't think I'm getting anywhere with this one lol

precal
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351479960238:dwgot this as well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the next few terms are something like (1/81/5+1/12)+(1/101/6+1/14)

precal
 4 years ago
Best ResponseYou've already chosen the best response.0I thought somehow we would be left with 1/4 as the first term then it would make sense that they solution is 9/4

precal
 4 years ago
Best ResponseYou've already chosen the best response.0wonder if we messed up on the partial fraction

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0Ummm I doubt it, it's likely that the terms just cancel out in a strange way :D

precal
 4 years ago
Best ResponseYou've already chosen the best response.0well that is good to know

precal
 4 years ago
Best ResponseYou've already chosen the best response.0yes, wonder if we have to do more than the first three terms to see it

precal
 4 years ago
Best ResponseYou've already chosen the best response.0ok that is all I see right now

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0nasty little problem :3

precal
 4 years ago
Best ResponseYou've already chosen the best response.0yes, well I think I will sleep on it, would not be the first time I went to bed with an unsolved problem....thanks..........

precal
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351531637965:dwok I think I solved it @zepdrix

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0Hmmmmmm that doesn't quite look right <:o you can't split up the base of a fraction. \[\frac{ 1 }{ 2n+4 }=\frac{ 1 }{ \infty+4 }=0\]