Question

Can anyone familiar with series convergence and divergence help me decipher this problem? it's poorly worded and i'm having trouble figuring out exactly what it is asking.

Answer 1

Let S be the sum of the numbers sin x over those x such that the series:
\[\sum_{n=0}^{infinity}(-1)^nn^3(x^3-x^2-8x+7)^n/(3n^4+1)\]
converges conditionally. The value of S is?

Answer 2

Or just, how do you find which x values cause it to converge conditionally.?

Answer 3

Thinking out loud here...You can determine if a series converges by finding the limit, much like you would a function, though I'm not certain how the sin(x) fits into it just yet.

Answer 4

yeah, the sinx part I think just forget about that. it's not really relevant to what the question is really asking. we need values for x such that the series converges conditionally. and that's kind of where I am stuck. How do you test for conditional convergence?

Answer 5

Keep in mind that this is exactly where I am currently in my Calc II so I'm not 100%, but the "conditional convergence" could be what "Paul" calls partial sum convergence: http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx

Answer 6

Now, since the main topic of this section is the convergence of a series we should mention a stronger type of convergence. A series is said to converge absolutely if also converges. Absolute convergence is stronger than convergence in the sense that a series that is absolutely convergent will also be convergent, but a series that is convergent may or may not be absolutely convergent.

In fact if converges and diverges the series is called conditionally convergent.

Answer 7

that's what he said in there about it. oh but it didn't copy the symbols.

Answer 8

the values for x on which it converges is called the interval of convergences i beleive

Answer 9

So this is part of the power series?

Answer 10

Right, and you're supposed to test the endpoints or something?

Answer 11

\[\lim_{n\to\ inf}\frac{(-1)^nn^3(x^3-x^2-8x+7)^n}{3n^4+1}*\frac{3(n-1)^4+1}{(-1)^{n-1}(n-1)^3(x^3-x^2-8x+7)^{n-1}}\]

Answer 12

\[\lim_{n\to\ inf}\frac{-n^3(x^3-x^2-8x+7)}{3n^4+1}*\frac{3(n-1)^4+1}{(n-1)^3}\]

whatever that simplifies to; we pull out the non-n parts

Answer 13

okay well if I were to say that i'm pretty sure that limit cancels out to
|-(x^3-x^2-8x+7)|, what do you suggest I can try next?

Answer 14

we need to know if the n parts limit to 1 or 0; if 0 then the intervals i -inf to inf

if 1 then the radius is from -1 to 1 if thats what we pulled out

Answer 15

so.. would I try and solve for x values like this.?

-1<(x^3-x^2-8x+7)<1

Answer 16

or can i just use -1 and 1.. or how does that work

Answer 17

im tryng to get the wolf to verify my idea without going to timedout mode:)

Answer 18

http://www.wolframalpha.com/input/?i=limit+%28-n%5E3%28%28x%5E3%E2%88%92x%5E2%E2%88%928x%2B7%29*3%28n-1%29%5E4%2B1%29%2F%28%283n%5E4%2B1%29%28n-1%29%5E3%29%2C+n+to+inf

well, your outcome is valid at least :)

Answer 19

Let S be the sum of the numbers sin x over those x such that the series:
So does this mean: S = sin(1)/1 + sin(2)/2 + sin(3)/3 ?

Answer 20

|-x^3+x^2+8x-7)| <= 1 such that 1 is the radius of convergence

-1 <= -x^3+x^2+8x-7) <= 1

is our interval of convergence

Answer 21

-x^3+x^2+8x-7-1 = 0

-x^3+x^2+8x-7+1 = 0

Answer 22

and then solving that .. would all the solutions for x allow for conditional convergence?
and is it less than or equal to? or just less than. I understood that at p = 1 the ratio test is inconclusive..

Answer 23

http://www.wolframalpha.com/input/?i=-x%5E3%2Bx%5E2%2B8x-7%3C%3D1+and+-x%5E3%2Bx%5E2%2B8x-7+%3E%3D+-1

there seem to be 3 interval what satisfy those conditions

Answer 24

So you are saying those endpoints of the three intervals (6 values for x) are the ones that the question is asking for.. the sin(x)/x sum?

Answer 25

im saying that that is what i come up with for your summation; whether your summation is correct or not, i dunno

Answer 26

quite frankly, i got no idea if my thought is even correct :/

Answer 27

Well, now I am just wondering whether maybe the answers are slightly off, because I did this exact same method already and got that exact same three intervals. And I really don't know how else to do it. but something is not working

Answer 28

is this your wording of the question? or is this the question itself?

Answer 29

That's the question itself..

Answer 30

horribly worded.

Answer 31

well then, if we could get it graphed we would be able to see how well it fits

Answer 32

it does appear to hit sinx/x in 3 spots

Answer 33

really? what graph are you using I have no idea how you're getting that?

Answer 34

i just used the first few sums of the summation you provided; and mapped it with sinx/x

Answer 35

hmm. so what values were they?

Answer 36

http://www.wolframalpha.com/input/?i=sin%28x%29%2Fx+%3D++-%28x%5E3%E2%88%92x%5E2%E2%88%928x%2B7%29%2F%283%2B1%29+%2B8%28x%5E3%E2%88%92x%5E2%E2%88%928x%2B7%29%5E2%2F%283%282%29%5E4%2B1%29+%E2%88%9227%28x%5E3%E2%88%92x%5E2%E2%88%928x%2B7%29%5E3%2F%283%283%29%5E4%2B1%29%2B4%5E3%28x%5E3%E2%88%92x%5E2%E2%88%928x%2B7%29%5E4%2F%283%284%29%5E4%2B1%29%2C+y%3D-10+to+10

they appear to match up with the interval that we found quite nicely

Answer 37

just using the first 4 sums; we hit in about -3, 1 and 3

Answer 38

Yeah, i'm not sure. I think we may be interpreting the sinx portion of the question incorrectly. I think it's just asking us to find the x's needed to make that series conditionally convergent. and then just take the sum of sin(x1)/x1+sin(x2/x2) ..etc. depending on how many x values satisfy conditional convergence.

Answer 39

sin(x2)/x2**

Answer 40

well, good luck with it :)

Answer 41

They usually add weird computational portions to questions for our assignments

Answer 42

yeah, thanks for your help.