sequence problem _ Calculus

- precal

sequence problem _ Calculus

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- precal

|dw:1351476871325:dw|

- precal

don't I need to use some formula?

- zepdrix

So is the question, "Does the series converge?" ? :o

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## More answers

- precal

yes, I know it converges to 9/4

- precal

doesn't help that I don't know why?

- precal

|dw:1351477345222:dw|

- precal

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- zepdrix

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Hmm can't you do some type of comparison test?
Sorry I'm a little rusty with these :D
We know that the p-series 1/n^2 converses, and our series is smaller than that one :O something like that..

- precal

|dw:1351477424396:dw|

- precal

I am as rusty as you

- zepdrix

Hmm :[

- precal

how about if I do this as a telescoping series? as if I know what I am talking about

- zepdrix

Yah that's prolly the way to approach it c: try to recognize a pattern coming out of it.

- precal

I know if I factor out the 9 and then I just need to create the (1/4)

- zepdrix

Oh 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

*

- zepdrix

|dw:1351478870156:dw|

- zepdrix

|dw:1351479010862:dw|

- anonymous

|dw:1351479486392:dw|

- zepdrix

oh sorry XD hehe

- precal

sorry, I ran off to watch some youtube videos on this

- zepdrix

Mmmm that's prolly a good idea :) I don't think I'm getting anywhere with this one lol

- precal

|dw:1351479960238:dw|got this as well

- precal

|dw:1351480090695:dw|

- anonymous

the next few terms are something like
(1/8-1/5+1/12)+(1/10-1/6+1/14)

- precal

I 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

wonder if we messed up on the partial fraction

- anonymous

I don't think so

- zepdrix

Ummm I doubt it, it's likely that the terms just cancel out in a strange way :D

- precal

well that is good to know

- precal

yes, wonder if we have to do more than the first three terms to see it

- zepdrix

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- precal

|dw:1351480632232:dw|

- zepdrix

Oh hmm :D

- precal

ok that is all I see right now

- zepdrix

nasty little problem :3

- precal

yes, well I think I will sleep on it, would not be the first time I went to bed with an unsolved problem....thanks..........

- precal

|dw:1351531637965:dw|ok I think I solved it @zepdrix

- zepdrix

Hmmmmmm that doesn't quite look right <:o you can't split up the base of a fraction.
\[\frac{ 1 }{ 2n+4 }=\frac{ 1 }{ \infty+4 }=0\]

- zepdrix

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