
#1
Dec1509, 06:15 PM

P: 26

1. The problem statement, all variables and given/known data
Given the infinite series, [tex]\sum[/tex] n=1 to [tex]\infty[/tex] of sin(nx)/n^(s) is convergent for 0 < s <= 1. 2. Relevant equations 3. The attempt at a solution Let f_n(x) = sin(nx) and g_n(x) = 1/n^(s) i.) lim n> [tex]\infty[/tex] f_n = 0 I'm not sure how to show this formally. Specifically, for a sequence instead of a function. ii.) [tex]\sum[/tex] g_(n+1)  g_n converges. Or this either. iii.) [tex]\sum[/tex] g_n, its partial sums are uniformly bounded. [tex]\sum[/tex] g_(n+1)  g_n = (g_1  g_2) + (g_2  g_3) + .... + (g_n  g_(n+1) = g_1  g_(n+1). Lim n>infinity [tex]\sum[/tex] g_(n+1)  g_n= lim n>infinity (g_1  g_(n+1)) = g_1. Therefore the partial sums are bounded, so [tex]\sum[/tex] n=1 to [tex]\infty[/tex] of sin(nx)/n^(s) is convergent for 0 < s <= 1. 



#2
Dec1609, 12:53 AM

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P: 20,962

Unless I am mistaken, this will be very difficult to prove, because it isn't true. [tex]S_k~=~\sum_{n = 1}^k \frac{1}{n^s} [/tex] 



#3
Dec1609, 07:13 AM

P: 26

Hi, thank you for replying. Here's a rewording of the problem.
Use the Dirichlet test to show that the infinite series from n=1 to infinity of sin(nx)/n^(s) is convergent for 0 less than s less than or equal to 1. Note: x is any real number. 



#4
Dec1609, 08:56 AM

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P: 20,962

Dirichlet Test
The Dirichlet test applies to series of the form
[tex]~\sum_{n = 1}^{\infty} a_nb_n [/tex] To use this test you need to show that the following are true:
The first two items are pretty easy to show, but the third requires some work. 



#5
Dec1609, 10:59 AM

P: 26

Alright, so,
So, i) Let (n+1)^(s) = n^s + sC1 x^(s1)(1) + sC2 x^(s2)(1)^2 + ..... Now, assuming n ≥ 0 we get that n^s is a small part or 1st term of the right hand side of above expression is ≥ the left hand side. Note: The other part sC1x^(s1)(1) + sC2 x^(s2)(1)^2 + ... is positive and subtracted to get n^s or n^s ≥ (n+1)^(s) Then, taking the reciprocal we change signs and rearranging we get 1/n^(s) ≥ 1/(n+1)^(s). ii) Let be n = 1+δ, δ>0. Let be S so that δ>1/S Therefore n^s > (1+1/S)^s = (S+1)^s/S^s = = (S^s + s S^(s1) + ...)/S^s > > s/S If s>SM, then n^s > M and then for each ε>0 there is M > 1/ε and for every s > S_o = SM, 1/n^s < 1/M < ε Then, lim 1/n^s = 0. iii) I don't really know where to start here, honestly. 



#6
Dec1609, 01:05 PM

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P: 20,962

No. Look at what I wrote in post #4. Steps 1 and 2 have to do with a_{n}, which I have identified as 1/n^{s}. Those steps are pretty easy to show.
Step 3 deals with b_{n}, which I am identifying as sin(nx). You have to show that there is a positive constant M for which the following inequality is true for all N and any x. [tex]\left\sum_{n = 1}^N sin(nx) \right \leq M[/tex] 



#7
Dec1609, 02:08 PM

P: 26

What did I do wrong in steps 1 & 2?




#8
Dec1609, 03:05 PM

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P: 20,962

I glossed over what you were doing and misunderstood what you were saying. In any case, to show steps 1 and 2 for a_n = 1/n^s, you can probably say what needs to be said in less than a quarter of what you said. You really don't need to expand (n + 1)^2, and you don't need to proved via epsilons and deltas that 1/n^2 > 0.




#9
Dec1609, 03:06 PM

P: 26

Alright, here's my attempt at part 3.
[tex]\sum[/tex] sin(nx) = [(cos (x/2)  cos(n + 1/2)x) / (2sin(x/2))] for all x with sin(/2) =/= 0. We have, sin(x/2) [tex]\sum[/tex] sin(nx) = sin(x/2) sinx + sin(x/2) sin 2x + ... + sin (x/2) sin(n) By the identity, 2 sin A sin B = cos (BA)  cos (B + A), this becomes, 2sin(x/2) [tex]\sum[/tex] sin(nx) = cos (x/2)  cos (n + 1/2)x. So, the partial sums are bounded by 1/sin(x/2). Therefore, [tex]\sum[/tex] sin(nx)/n^(s) converges. 



#10
Dec1609, 04:33 PM

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P: 20,962

This looks like a good start at it. Some comments along the way.
I think this is what you mean to say. [tex]\sum_{n = 1}^N sin(nx)~=~\frac{cos (x/2)  cos((N + 1/2)x)}{2sin(x/2)} [/tex] for all x such that sin(x/2) =!= 0. [tex]\sum_{n = 1}^N sin(nx)~=~\frac{cos (x/2)  cos((N + 1/2)x)}{2sin(x/2)} [/tex] so how do you conclude that this partial sum is bounded by 1/sin(x/2)? 



#11
Dec1609, 04:51 PM

P: 26

I left out some work, let me pick back up here:
By the identity, 2 sin A sin B = cos (BA)  cos (B + A), this becomes, 2sin(x/2)[tex]\sum[/tex] from 1 to n of sin kx = (cos x/2  cos 3x/2) + (cos 3x/2 + cos 5x/2) + ... + (cos (n1/2)x  cos (n+1/2)x) = cos (x/2)  cos (n + 1/2)x. 



#12
Dec1609, 06:16 PM

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P: 20,962

I understand that. My question is how do you conclude that this partial sum is bounded by 1/sin(x/2)?




#13
Dec1609, 07:34 PM

P: 26

Wait, lets go back to the second part here.
a_n >= a_n+1 > 0, with an = 1/ns Shouldn't that be ∑a_(n+1)  a_(n) converges instead?? 



#14
Dec1609, 08:01 PM

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P: 20,962

No, not at all. To use the Dirichlet test (which is used on a summation of the product of elements of two sequences), one of the sequences has to be decreasing (a_n >= a_n+1) and bounded below by zero. That's the same as saying this sequence converges to zero. Where are you getting this: ∑a_(n+1)  a_(n) ?




#15
Dec1609, 08:49 PM

P: 50

I got it from my textbook, actually, haha.




#16
Dec1609, 11:52 PM

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Thanks
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#17
Dec1709, 12:20 AM

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P: 20,962

That works for me. Thanks for jumping in, Dick. I don't feel intruded on at all.



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