# show series [sin(n)]/n converges?

by erjkism
Tags: converges, series, sinn or n
 HW Helper P: 2,080 Do you know how to find fourier expansions? consider (pi-x)/2 0
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Thanks
P: 24,460
 Quote by lurflurf 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.
 HW Helper P: 3,353 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 realised I found the wrong Fourier series. argh
 HW Helper Sci Advisor Thanks P: 24,460 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.
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Thanks
P: 24,460
 Quote by 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 realised 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.
P: 291
 Quote by lurflurf 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.
 HW Helper Sci Advisor Thanks P: 24,460 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.
P: 352
 Quote by 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.
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?
 HW Helper Sci Advisor Thanks P: 24,460 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.
P: 352
 Quote by 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.
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.
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 HW Helper Sci Advisor P: 2,533 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: $$\sum \frac{1}{n^2+n} f(n)$$
 P: 20 $$\sum_{n=1}^{\infty} a_n \, sin(n)$$ converges whenever {an} is a decreasing sequence that tends to zero. By : Dirichlet's test