1. The problem statement, all variables and given/known data [tex]\Sigma\frac{sin(n)}{n}[/tex] from n=1 to infinity i have to show that this series converges, but that it doesnt converge absolutely. 2. Relevant equations the thing is, I figured th 3. The attempt at a solution if you say that the top of the equation is -1<sin(n)<1, you can then take the absolute value to get 0<sin(n)<1. so then the equation can look like this: [tex]\Sigma\frac{1}{n}[/tex] from n=1 to infinity, which is a divergent series. i just dont know how to show that the original equation is convergent.
Try to use D'alemberts test for convergence of series. [tex]\lim_{n\to\infty}\frac{a_n_+_1}{a_n}=L[/tex] if L<1 it converges if L>1 it diverges if L=1 this criterion does not give any answer.
okay so i tried that, but i'm having a tough time figuring out the result of that limit. [tex]\frac{n sin(n+1)}{(n+1)sin(n)}[/tex] I am still stuck with a bunch of sin functions... im not sure if this is the right thing to do
Keep in mind that sin(x) never goes above 1 or below -1, but oscillates infinitely often as x approaches infinity. So you are really left with: [tex]\frac{n}{n+1}[/tex] and as n approaches infinity you can drop the constants and your limit is equal to 1, making the test inconclusive. Maybe try using the fact that [tex]\sum\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} = sin(x)[/tex] n=0 -> infinity from taylor series expansion of sin(x). Plugging that in you get a double summation that will oscillate between -1 and 1 for all values of x, I think.
The series doesn't alternate in any sort of reasonable pattern which I think puts this outside of the set of elementary problems. What course are you taking?
its just a problem that i need to do for calculus 2 at college... so i don't know much more than what's been talked about here already
Compare it to its integral. It's quite famous, its called the Dirchlet integral. Take a look at its proof of conditional convergence on the net, and if you think it's too advanced, either try another method or skip this question, or if you think the proof is reasonable, then its good =]
According to Wikipedia, the Dirichlet Integral is the integral of the same function as has been discussed, but evaluated from zero to infinity instead of the one to infinity to be considered here. Surely evaluation at zero and one would make a difference here since we are dealing with a trig. function? I've tried applying the Integral Test to this and it concludes divergence.
Try the result on wikipedia. Thats more of a proof of its exact value though, more than what you need.
You CANNOT do that. The integral test is only strictly valid for monotone functions. I think this problem landed in a calc 2 course by mistake.
no, our teacher told us that we have the ability to do it, but it is just pretty difficult. he wont help me... someone save me!!!
Ok, then maybe it's not so godawful hard. Here's some things to try. You can sum a series like r^n/n by differentiating with respect to r and obtaining a geometric series. sin(n)=Im(exp(in)). So your series can be written as Im(exp(i)^n/n). Try that and see where it takes you. I'll check back later and see how you are doing... It's debatable whether this is a 'proof' of convergence though. You're assuming the limit exists to begin with.
If you can do following? -1/x <sin(x)/x <1/x 1/x and -1/x converge(?) so does sin(x)/x? I don't think anyone has suggested this; maybe because I skimmed the thread too fast or this is wrong.
Dick, I'm not sure what you're doing up in your previous post, though it may just be the notation... Anyways, I'm in ap calc bc, which is pretty much like calc 2 I guess. I had to get kind of "creative" with it so I don't know if everything I did is exactly legit. So here's my answer and reasoning: (I've omitted the sigma notation for ease here) sin(n)/n is + when n:(0,pi) and - when n:(pi,2*pi) therefor the sign changes every pi. sin(n)/n already looks a lot like (-1)^n/n so I tried to figure out a way to get (-1)^n to alternate like sin(n) does. My first impulse was to use (-1)^(n/pi) but that leads to a problem since that would give you odd/even values for the exponent alternating infinitely fast. So, you've got to keep that exponent restricted to integers. This brings me to the floor() function (this is my "creative" part). I've never used this in formal math, but its pretty similar to [|x|] in that it rounds down the nearest integer which is just what I need to deal with that icky exponent. I also checked a few numbers really quick to make sure that my (-1)^floor(n/pi) had the same sign at sin(n) for various n's so I could actually compare them without getting into more problems. So, that gives me (let's pretend "E" is the equivalent to a sigma) n=1 to inf. E(-1)^floor(n/pi) This converges by the alternating series test. Which I spare us all from reading (and me from typing!), I'm sure you know how it goes. So we know n=1 to inf. E(-1)^floor(n/pi) converges and we want to show that n=1 to inf. Esin(n)/n converges. Direct comparison isn't an option since the terms have to be greater than zero so I went with the limit comparison test. sin(n)/n * n/((-1)^floor(n/pi)) = sin(n)/((-1)^floor(n/pi)) this result is a constant number. It will always be positive since the alternating signs of the two functions always end up canceling each other out in the series since n is an integer. Also the resulting ratio will never equal zero as n is restricted to integers so there is no way to get an irrational number like pi (and you skip zero since n starts at one) which would result in the sin(n) = 0. since you get a number greater than zero from the limit comparison test by comparing it with a convergent series the series n=1 to inf. Esin(n)/n converges! ...that was so much shorter on paper...
Well, that does show some creative thinking. And I think you know what some of the problems here are. But, I don't believe you. Because the sum of (-1)^floor(n/pi) DEFINITELY does not converge in any sense. It is not alternating for one thing, but it's biggest defect is that the terms don't even approach zero. No such thing can converge. What I'm suggesting is to use the Euler formula to express sin(n) as a the imaginary part of the power of a complex exponential. This makes the problem look closely related to a geometric series of dubious convergence. I'm kind of waiting to the OP to explore this before getting too specific.
Wont. [itex]\sum_{n=0}^{-\infty}\frac{sin(n)}{n}\;dn \rightarrow -\frac{1}{2}\pi[/itex] [itex]\sum_{n=0}^{\infty}\frac{sin(n)}{n}\;dn \rightarrow \frac{1}{2}\pi[/itex] [itex]\int_{-\infty}^{\infty}\frac{sin(n)}{n}\;dn -\int_{-1}^{1}\frac{sin(n)}{n}=2si(1)\rightarrow=\pi-2si(1)[/itex] Where si is the sigmoid function. So or putting it in terms of convergence the top divided by the bottom as the limit approaches infinty "converges" to [itex]\pi[/itex]-2si(1) Because 1/2 of that = 1/2[itex]\pi[/itex] and 1/2 of the other =[itex]-\pi[/itex]? so because of that we get the unusual result: [tex]\frac{\frac{sin(n)} {\infty}} {{n}{\infty}}=\pm\frac{1}{2}\pi \lim_{n\rightarrow\pm\infty}-2si(1)[/tex] That's meant to be [itex]\frac{n}{\infty}[/itex] in the denominator but for some reason latex wont do it. Actually that's a really weird way of putting it but it should converge to pi given that if we accept that -2(si) is a definite result or that +/- infinity- that is in fact pi. In other words no it does not converge unless we accept that it must have a solution that doesn't involve a non elementary function. If we don't then there is no solution, it does not converge.
The latex in that isn't correct, and I can no longer edit it. But I think you get the point? Or have I just confused you and myself? Should be: [tex] \frac{\frac{sin(n)} {\infty}} {n/\infty}=\pi \lim_{n\rightarrow\pm\infty}-2si(1) [/tex] [itex] \int_{-\infty}^{\infty}\frac{sin(n)}{n}\;dn \rightarrow\int_{-1}^{1}\frac{sin(n)}{n}=2si(1)\rightarrow=\pi-2si(1) [/itex] Or [itex]\pi-\frac{dP}{dt}=P(1-P).[/itex] Or [itex]\pi[/itex] - http://en.wikipedia.org/wiki/Image:Logistic-curve.png I bet that makes no sense whatsoever.