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Dirichlet Test

  1. 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. jcsd
  3. Mark44

    Staff: Mentor

    Can you give us the problem verbatim? What you have written is not as clear as it should be. Are you supposed to prove that your series converges if 0 < s <= 1?

    Unless I am mistaken, this will be very difficult to prove, because it isn't true.
    This is not true. For an arbitrary value of x, the sequence f_n(x) does not converge. The values lie in a band between -1 and 1.
    This might be true, but it doesn't have anything to do with this problem that I can see.
    The partial sums would be the sequence
    [tex]S_k~=~\sum_{n = 1}^k \frac{1}{n^s} [/tex]
     
  4. Hi, thank you for replying. Here's a re-wording 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.
     
  5. Mark44

    Staff: Mentor

    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:
    1. an >= an+1 > 0, with an = 1/ns (The other sequence, sin(nx), is not a decreasing, bounded sequence.)
    2. lim an = 0, as n --> [itex]\infty[/itex]
    3. [tex]\left|\sum_{n = 1}^N b_n \right| \leq M~for~all~N[/tex], and where M is a positive constant

    The first two items are pretty easy to show, but the third requires some work.
     
  6. Alright, so,

    So,

    i) Let (n+1)^(s) = n^s + sC1 x^(s-1)(1) + sC2 x^(s-2)(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^(s-1)(1) + sC2 x^(s-2)(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^(s-1) + ...)/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.
     
  7. Mark44

    Staff: Mentor

    No. Look at what I wrote in post #4. Steps 1 and 2 have to do with an, which I have identified as 1/ns. Those steps are pretty easy to show.

    Step 3 deals with bn, 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]
     
  8. What did I do wrong in steps 1 & 2?
     
  9. Mark44

    Staff: Mentor

    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.
     
  10. 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 (B-A) - 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.
     
  11. Mark44

    Staff: Mentor

    This looks like a good start at it. Some comments along the way.
    What you have below is hard to follow, since it appears to be the conclusion from the work below it. You should give the reader a clue as to where it comes from.

    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.
    Instead of saying "We have," you really mean "because."
    Now you've lost me. You have
    [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)|?
     
  12. I left out some work, let me pick back up here:

    By the identity, 2 sin A sin B = cos (B-A) - 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 (n-1/2)x - cos (n+1/2)x)

    = cos (x/2) - cos (n + 1/2)x.
     
  13. Mark44

    Staff: Mentor

    I understand that. My question is how do you conclude that this partial sum is bounded by 1/|sin(x/2)|?
     
  14. 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??
     
  15. Mark44

    Staff: Mentor

    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)| ?
     
  16. I got it from my textbook, actually, haha.
     
  17. Dick

    Dick 25,832
    Science Advisor
    Homework Helper

    Not to intrude, but the numerator is bounded by 2, right? I think it is bounded by 1/|sin(x/2)|.
     
  18. Mark44

    Staff: Mentor

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