Dirichlet Test

hth

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.

Mark44

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]

hth

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.

Mark44

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. a_n >= a_n+1 > 0, with a_n = 1/n^s (The other sequence, sin(nx), is not a decreasing, bounded sequence.)
2. lim a_n = 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.

hth

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.

Mark44

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]

hth

What did I do wrong in steps 1 & 2?

Mark44

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.

hth

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.

Mark44

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)|?

hth

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.

Mark44

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

hth

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??

Mark44

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)| ?

tracedinair

I got it from my textbook, actually, haha.

Dick

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

Mark44

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