Show series [sin(n)]/n converges?

[erjkism]

Homework Statement

\Sigma\frac{sin(n)}{n}

from n=1 to infinity


i have to show that this series converges, but that it doesn't converge absolutely.

Homework Equations

the thing is, I figured th

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:

\Sigma\frac{1}{n} from n=1 to infinity, which is a divergent series. i just don't know how to show that the original equation is convergent.

 

[sutupidmath]

Try to use D'alemberts test for convergence of series.

\lim_{n\to\infty}\frac{a_n_+_1}{a_n}=L if L<1 it converges if L>1 it diverges if L=1 this criterion does not give any answer.

 

[erjkism]

okay so i tried that, but I'm having a tough time figuring out the result of that limit.

\frac{n sin(n+1)}{(n+1)sin(n)}


I am still stuck with a bunch of sin functions... I am not sure if this is the right thing to do

 

[sutupidmath]

okay so i tried that, but I'm having a tough time figuring out the result of that limit.

\frac{n sin(n+1)}{(n+1)sin(n)}


I am still stuck with a bunch of sin functions... I am not sure if this is the right thing to do

it seems it doesn't work after all!

 

[exk]

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:
\frac{n}{n+1}
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 \sum\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} = sin(x) 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.

 

[Dick]

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?

 

[erjkism]

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

 

[Gib Z]

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 =]

 

[White Ink]

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.

 

[Gib Z]

Try the result on wikipedia. Thats more of a proof of its exact value though, more than what you need.

 

[Dick]

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 =]

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.

 

[erjkism]

no, our teacher told us that we have the ability to do it, but it is just pretty difficult. he won't help me... someone save me!

 

[Dick]

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.

 

[rootX]

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.

 

[rostbrot]

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...

 

[rostbrot]

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.

neither -1/n nor 1/n converge.
(-1)^n/n does, however, converge.

 

[rootX]

Thanks ;)
I thought sequences should be little different from series.

 

[Dick]

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.

 

[Schrodinger's Dog]

Wont.

\sum_{n=0}^{-\infty}\frac{sin(n)}{n}\;dn \rightarrow -\frac{1}{2}\pi

\sum_{n=0}^{\infty}\frac{sin(n)}{n}\;dn \rightarrow \frac{1}{2}\pi

\int_{-\infty}^{\infty}\frac{sin(n)}{n}\;dn -\int_{-1}^{1}\frac{sin(n)}{n}=2si(1)\rightarrow=\pi-2si(1)

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 \pi-2si(1)

Because

1/2 of that = 1/2\pi and 1/2 of the other =-\pi?

so because of that we get the unusual result:

\frac{\frac{sin(n)} {\infty}} {{n}{\infty}}=\pm\frac{1}{2}\pi \lim_{n\rightarrow\pm\infty}-2si(1)

That's meant to be \frac{n}{\infty} in the denominator but for some reason latex won't 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.

 

[Schrodinger's Dog]

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:

<br /> \frac{\frac{sin(n)} {\infty}} {n/\infty}=\pi \lim_{n\rightarrow\pm\infty}-2si(1)<br />

<br /> \int_{-\infty}^{\infty}\frac{sin(n)}{n}\;dn \rightarrow\int_{-1}^{1}\frac{sin(n)}{n}=2si(1)\rightarrow=\pi-2si(1)<br />

Or \pi-\frac{dP}{dt}=P(1-P).

Or \pi -

http://en.wikipedia.org/wiki/Image:Logistic-curve.png

I bet that makes no sense whatsoever.

 

[Schrodinger's Dog]

<br /> <br /> 1. \int_{-\infty}^{\infty}\frac{sin(n)}{n}\;dn \rightarrow\int_{-1}^{1}\frac{sin(n)}{n}=2si(1)\rightarrow=\pi-2si(1)\; \text{given eq 1.}+2si(1)=\pi \;or \; \text{or does not converge}<br /> <br />

And that should be so it converges to pi or does not converge according to how you want to float your boat.

 

[Dick]

<br /> <br /> 1. \int_{-\infty}^{\infty}\frac{sin(n)}{n}\;dn \rightarrow\int_{-1}^{1}\frac{sin(n)}{n}=2si(1)\rightarrow=\pi-2si(1)\; \text{given eq 1.}+2si(1)=\pi \;or \; \text{or does not converge}<br /> <br />

And that should be so it converges to pi or does not converge according to how you want to float your boat.

? This problem has NOTHING to do with the sine integral, Si. Nor is the sum pi/2. I'll give you all a big hint. I summed 1000000 terms of the series. I get about 1.0708. It converges pretty slowly like a harmonic series. I also know a combination of elementary functions that gives me that number. I found this by following my own suggestion. So I believe it's correct. 1.0708 won't tell you what the answer is, but it will tell you when your answer is wrong. And 1.0708 is NOT pi/2.

 

[Schrodinger's Dog]

Actually, I was talking about the integral of that.

as in between the range of -\infty to infty.

And between 1 and infinity and 1 and 1.

if the integral is this then in sum form it is this.

Then we can work out if the integral of sin(n)/n converges? Does that make more sense?

I was trying to prove that in fact the integral of 1 to infinity that equation is 1/2 pi because the total integral from negative infinity to infinity is pi-2si(1).

I was wondering if that is a fair assumption, I can see how it might of looked like I was trying to answer the op though, my bad. What I should of done is prefaced it with say if we have the integral here, how do I prove that it does or does not converge?

Can we prove that it is equal to pi on its own or to pi-si(2) or that there is nos substantive difference between the two? Does that make sense, if not, sorry. I was just thinking aloud you know?

As for the sequence well that's another matter. Sorry. And one that I think have enough clues there to solve it now.

Unfortunately I cannot edit any of my posts so I'm stuck with explaining it here.

Obviously if I type the summation into a program it gives me a pretty good answer, so I'll not go there.

 

[Dick]

If you are just messing around with the integral knowing it has nothing to do with the series, I'm not sure this is the place to do that. It's going to confuse the OP.

 

[Schrodinger's Dog]

If you are just messing around with the integral knowing it has nothing to do with the series, I'm not sure this is the place to do that. It's going to confuse the OP.

K my bad. Is it fair to say that there is no substantial or significant difference between the answer pi and pi-si(2) though? Is what I was asking? Or would that be an erroneous assumption?

I'll delete those posts if I can. And this one if you like pm me. or spoiler an answer.

 

[Dick]

It's pretty obvious there there is a difference between pi and pi-si(2). For one thing, they are different numbers. I don't know what your question is. Can you compose it in a different thread? I'd be happy to look at it there.

 

[Schrodinger's Dog]

It's pretty obvious there there is a difference between pi and pi-si(2). For one thing, they are different numbers. I don't know what your question is. Can you compose it in a different thread? I'd be happy to look at it there.

Sure. It was only the idea that -Si(1) and -Si(1) may cancel given an infinite spectrum and thinking pictorially about the wave, ie if we look at it terms of an overall number. But I think I can see why that is wrong now, ie they function is not a mirror function either side of 0, so never mind. It was just an idle speculation, apologies for the thread jack, I've asked for the posts to be deleted so as to avoid confusion.

 

[Gib Z]

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.

My bad :( Sorry for misleading the OP !

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.

I very much like that solution :D Excellent work Dick =]

 

[DavidWhitbeck]

None of the solutions are acceptable yet considering the student is in AP Calculus. And sadly that includes using DeMoivre's formula (that's right many students that you think would know that formula by AP Calc don't!). There must be a solution only using the standard tests that they are taught (limit comparison, ratio, root, integral, etc), and it should be just a few lines long.

 

[Pere Callahan]

That's a very interesting question. I agree Dick's method is very nice and seems to work. However I also agree that the techniqes used might be too advanced.
A proof using only elementary comparison tests and the like would certainly be much more technical and I see no reason why it should be only a few lines long. I did something similar some time ago maybe I can find my notes.

 

[DavidWhitbeck]

That's a very interesting question. I agree Dick's method is very nice and seems to work. However I also agree that the techniqes used might be too advanced.
A proof using only elementary comparison tests and the like would certainly be much more technical and I see no reason why it should be only a few lines long. I did something similar some time ago maybe I can find my notes.

I said that it should be only a few lines long because otherwise it's not an acceptable problem for a high school calculus class.

And the reason it should involve comparison tests is because that's what they're taught in the class (standard topic for Calc BC which the OP said was taking). The problem must certainly be an exercise in the tools that they were taught. In a class like that, they're not going to deal with fun problems that require tricks, which is a pity.

Also comparison tests make up the last topic in a BC course, and it's April, if a student is posting a question now it's either from reviewing AP exams or because they are still studying comparison tests!

 

[Dick]

That's a very interesting question. I agree Dick's method is very nice and seems to work. However I also agree that the techniqes used might be too advanced.
A proof using only elementary comparison tests and the like would certainly be much more technical and I see no reason why it should be only a few lines long. I did something similar some time ago maybe I can find my notes.

I agree with you guys. My 'method' isn't even a proof. It involves manipulating divergent series. I think a genuine proof would involve real analysis techniques, e.g. you are definitely going to need that pi is irrational, and that's not an elementary proof. When the instructor said the problem was difficult but they could do it the only thing simple enough I could think of was deMoivre. I'd LOVE to see this elementary proof. Find those notes Pere Callahan!

 

[DavidWhitbeck]

Yeah find those notes! I'm going to ask a couple of people that would know better than me-- my students. They are taking Calc BC now, and they get 5's on every test. Since it's fresh on their minds I bet they could figure it out real fast, while I'm still stuck scratching my head trying to remember what it was all about.

 

[lurflurf]

try sumation by parts
ΣuΔv=uv-ΣΔuEv
btw the sum is (π-1)/2

 

[drpizza]

How about writing the numerator as a MacLaurin Series, then dividing out the n in the denominator? Then, use a comparison test and compare it to the MacLaurin Series for Cos(n)?

\sin{n} = n - \frac{n^3}{3!} +\frac{n^5}{5!} - \frac{n^7}{7!} +\frac{n^9}{9!} . . .

Therefore,

\Sigma\frac{\sin{n}}{n} = 1 -\frac{n^2}{3!} +\frac{n^4}{5!} - \frac{n^6}{7!} +\frac{n^8}{9!} . . .

Compare that to

\cos{n} = 1 - \frac{n^2}{2!} +\frac{n^4}{4!} - \frac{n^6}{6!} +\frac{n^8}{8!} . . .

 

[lurflurf]

How about writing the numerator as a MacLaurin Series, then dividing out the n in the denominator? Then, use a comparison test and compare it to the MacLaurin Series for Cos(n)?

Oh my you have lost the sum. Thus you have introduced a double sum. Analysis of such a some is likely harder than that of the original sum.

 

[lurflurf]

Do you know how to find Fourier expansions?
consider
(pi-x)/2 0<x<2pi
then consider x=1

 

[Dick]

try sumation by parts
ΣuΔv=uv-ΣΔuEv
btw the sum is (π-1)/2

Hi lurflurf,

You obviously know what you are doing. I was writing the answer as arctan(sin(1)/(1-cos(1))), but yeah, that is more cleanly expressed as (pi-1)/2. But how do you set this up using this 'summation by parts'? And I haven't figured out the how the Fourier expansion works yet, but I'll look at it tomorrow. Really nice work. Most other contributions to this thread have been completely clueless.

 

[Gib Z]

I haven't ever seen this summation by parts formula before, though I should have expected it exists, there must have been some discrete form for the continuous counterpart. As for the Fourier expansion, that seems even more out of such a course, though It is also a nice solution =]

Dick; as for the Fourier expansion, since f(x) = x is an odd function, the a coefficient is automatically zero whilst b_v = \frac{2}{\pi} \int^{\pi}_0 x \sin vx dx which some quick integration by parts yields b_v = (-1)^{v+1} \frac{2}{v}, and so we get x = 2 \left( \frac{\sin x}{1} - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + ... \right).

Rearranging and letting x=1 gives us the desired result, and putting in x= pi/2 gives us a very famous result =]

EDIT: O god i just realized I found the wrong Fourier series. argh

 

[Dick]

Hmm. If you did everything right then x=1 does not give the desired solution. That's an alternating sign series (-1)^n*sin(n)/n. The original question doesn't have the alternating sign. Is there an extra sign that makes it work? But like I said, I didn't really look at that yet. I was trying to figure out the summation by parts thing. It makes perfect sense it such a thing would exist though. Thanks, Gib.

 

[Dick]

I haven't ever seen this summation by parts formula before, though I should have expected it exists, there must have been some discrete form for the continuous counterpart. As for the Fourier expansion, that seems even more out of such a course, though It is also a nice solution =]

Dick; as for the Fourier expansion, since f(x) = x is an odd function, the a coefficient is automatically zero whilst b_v = \frac{2}{\pi} \int^{\pi}_0 x \sin vx dx which some quick integration by parts yields b_v = (-1)^{v+1} \frac{2}{v}, and so we get x = 2 \left( \frac{\sin x}{1} - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + ... \right).

Rearranging and letting x=1 gives us the desired result, and putting in x= pi/2 gives us a very famous result =]

EDIT: O god i just realized I found the wrong Fourier series. argh

Ok, if you do the right Fourier series, it does work. And yep, putting in x=pi/2 gives you the Gregory-Leibniz formula for pi. Interesting.

 

[drpizza]

Oh my you have lost the sum. Thus you have introduced a double sum. Analysis of such a some is likely harder than that of the original sum.

Ooops! That was as silly mistake. Should have worked it on paper first - I lose things when I'm typing them sometimes.

 

[Dick]

For future reference, a systematic approach to this uses the Abel-Dedekind-Dirichlet theorem. The product of a sequence with bounded partial sums (sin(n), use a trig identity) and a function of bounded variation (1/n), converges as a series. This is proved using Abel's 'summation by parts', as lurflurf intimated.

 

[DavidWhitbeck]

For future reference, a systematic approach to this uses the Abel-Dedekind-Dirichlet theorem. The product of a sequence with bounded partial sums (sin(n), use a trig identity) and a function of bounded variation (1/n), converges as a series. This is proved using Abel's 'summation by parts', as lurflurf intimated.

I came across that theorem yesterday and I was seeing if it fit with that problem, but I don't think it does because the partial sums of sin(n) must be bounded. How does one show that those partial sums are bounded?

 

[Dick]

Use sin(nx)=[cos((n-1/2)x)-cos(n+1/2)x)]/(2*sin(x/2)). It's a telescoping series. So if A_n is the partial sum, |A_n|<=1/|sin(x/2)|. Clever, huh? Wish I'd thought of it. It shows things like sin(2n)/n converge also.

 

[DavidWhitbeck]

Use sin(nx)=[cos((n-1/2)x)-cos(n+1/2)x)]/(2*sin(x/2)). It's a telescoping series. So if A_n is the partial sum, |A_n|<=1/|sin(x/2)|. Clever, huh? Wish I'd thought of it. It shows things like sin(2n)/n converge also.

That is very clever!

Well that problem would be appropriate for upper level math (analysis for use of the Dirichlet test, or applied math for engineers/physicists for recognizing that it's a Fourier series). But I think it's now safe to say that was probably an inappropriate problem for freshman calculus.

 

[Dick]

That is very clever!

Well that problem would be appropriate for upper level math (analysis for use of the Dirichlet test, or applied math for engineers/physicists for recognizing that it's a Fourier series). But I think it's now safe to say that was probably an inappropriate problem for freshman calculus.

Probably so, I think the student who asked this ran away long ago.

 

[NateTG]

When I look at something like that, my temptation is to try to find an upper bound for:
f(n)=\sum_{j=1}^{k} \sin(jx)
using geometry, and then to rearrange things into a sum that should look like:
\sum \frac{1}{n^2+n} f(n)

 

[Ratio Test =)]

\sum_{n=1}^{\infty} a_n \, sin(n)
converges whenever {an} is a decreasing sequence that tends to zero.

By : "[URL Black"]Dirichlet's test[/B][/U][/URL]

 

[myriadbeings]

Dirichlet's test will work for this series.
we can think of sin(n)/n as the product sin(n)*1/n.
dirichlet's test says that if one of the sequences in the product is bounded, and the other is monotone and converges to 0 then the series of the product of sequences must converge.
the sequence of partial sums of sin(n) is bounded and certainly 1/n is monotone and converges to 0, so the series must converge.