Proving the Convergence of \sum n=1 to \infty sin(nx)/n^(s)

In summary: You want to say that for any x such that sin(x/2) != 0, because of the identity you just mentioned, the sum in the previous line is bounded by\sum_{n = 1}^N sin(nx)~=~\frac{cos (x/2) - cos((N... Do you see what I mean?I can't make sense of what you've written. After the 2sin(x/2) \sum sin(nx) = cos (x/
  • #1
hth
26
0

Homework Statement



Given the infinite series, [tex]\sum[/tex] n=1 to [tex]\infty[/tex] of sin(nx)/n^(s) is convergent for 0 < s <= 1.

Homework Equations



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.
 
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  • #2
hth said:

Homework Statement



Given the infinite series, [tex]\sum[/tex] n=1 to [tex]\infty[/tex] of sin(nx)/n^(s) is convergent for 0 < s <= 1.
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.
hth said:

Homework Equations



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
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.
hth said:
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.
This might be true, but it doesn't have anything to do with this problem that I can see.
hth said:
Or this either.

iii.) [tex]\sum[/tex] g_n, its partial sums are uniformly bounded.
The partial sums would be the sequence
[tex]S_k~=~\sum_{n = 1}^k \frac{1}{n^s} [/tex]
hth said:
[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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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]
 
  • #7
What did I do wrong in steps 1 & 2?
 
  • #8
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
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.
 
  • #10
This looks like a good start at it. Some comments along the way.
hth said:
Alright, here's my attempt at part 3.
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.
hth said:
[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,
Instead of saying "We have," you really mean "because."
hth said:
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.
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)|?
 
  • #11
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.
 
  • #12
I understand that. My question is how do you conclude that this partial sum is bounded by 1/|sin(x/2)|?
 
  • #13
Wait, let's 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
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
I got it from my textbook, actually, haha.
 
  • #16
Mark44 said:
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)|?

Not to intrude, but the numerator is bounded by 2, right? I think it is bounded by 1/|sin(x/2)|.
 
  • #17
That works for me. Thanks for jumping in, Dick. I don't feel intruded on at all.
 

Related to Proving the Convergence of \sum n=1 to \infty sin(nx)/n^(s)

1. What is the purpose of proving the convergence of a series?

The purpose of proving the convergence of a series is to determine whether the series will eventually reach a finite sum or continue to increase without bound. This is important in mathematical analysis and applications in physics and engineering.

2. How do you prove the convergence of a series?

To prove the convergence of a series, several methods can be used such as the comparison test, ratio test, root test, integral test, and limit comparison test. These methods involve analyzing the behavior of the terms in the series and determining if they meet certain conditions for convergence.

3. What is the significance of the variable "s" in the series?

The variable "s" in the series represents the power to which the denominator is raised. This power can greatly affect the convergence or divergence of the series, and finding the specific value of "s" can be crucial in determining the convergence of the series.

4. Can the convergence of a series be proven for all values of "s"?

No, the convergence of a series can only be proven for certain values of "s". Depending on the specific series and method used to prove convergence, there may be a range of values for "s" that result in convergence. For example, the series in question converges for "s>1" and diverges for "s<=1".

5. Are there any real-world applications for proving the convergence of a series?

Yes, there are many real-world applications for proving the convergence of a series. For example, in physics and engineering, knowing the convergence of a series can help in predicting the behavior of systems over time. It can also be used in financial analysis to determine the growth or decline of investments over time.

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