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Finding the r value for the series to converge

  • #1
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Homework Statement


Let ##r\in \mathbb{R}##. Determine whether ##\displaystyle \sum_{n=1}^{\infty}\sin \left(\frac{n\pi}{3}\right)\frac{1}{n^r}## converges

Homework Equations




The Attempt at a Solution


If ##r>1##, then ##|\sin \left(\frac{n\pi}{3}\right)\frac{1}{n^r}| \le |\frac{1}{n^r}|##. The latter converges, so by the direct comparison test, the original series converges.

Now I want to show that when ##-\infty < r \le 1## the series diverges, but I am having trouble figuring out how. Should I show that the limit of the terms is not zero?
 

Answers and Replies

  • #2
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Homework Statement


Let ##r\in \mathbb{R}##. Determine whether ##\displaystyle \sum_{n=1}^{\infty}\sin \left(\frac{n\pi}{3}\right)\frac{1}{n^r}## converges

Homework Equations




The Attempt at a Solution


If ##r>1##, then ##|\sin \left(\frac{n\pi}{3}\right)\frac{1}{n^r}| \le |\frac{1}{n^r}|##. The latter converges, so by the direct comparison test, the original series converges.

Now I want to show that when ##-\infty < r \le 1## the series diverges, but I am having trouble figuring out how. Should I show that the limit of the terms is not zero?
If ##r \in (-\infty, 0]##, it's easy to show that the series diverges, using the Nth term test for divergence (i.e,, if ##\lim_{n \to \infty}a_n \ne 0##, then ##\sum a_n## diverges). For ##r \in (0, 1]##, a comparison of limit comparison test might be used.
 
  • #3
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If ##r \in (-\infty, 0]##, it's easy to show that the series diverges, using the Nth term test for divergence (i.e,, if ##\lim_{n \to \infty}a_n \ne 0##, then ##\sum a_n## diverges). For ##r \in (0, 1]##, a comparison of limit comparison test might be used.
I'm actually having a hard time showing the case when ##r\in(-\infty, 0]##. For argument's sake let ##p=-r##, then we have ##\lim \sin(\frac{n\pi}{3})n^p##. But I am not sure how to explicitly show that this product diverges. I am not even sure how to show that ##\sin(\frac{n\pi}{3})## diverges...
 
  • #4
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I'm actually having a hard time showing the case when ##r\in(-\infty, 0]##. For argument's sake let ##p=-r##, then we have ##\lim \sin(\frac{n\pi}{3})n^p##. But I am not sure how to explicitly show that this product diverges.
Let's look at things with some numbers, say r = -1.
##\frac 1 {n^r} = \frac 1 {n^{-1}} = n##. What is ##\lim_{n \to \infty} \frac 1 {n^r}##?
What if r = -1/2? What if r = -2? Etc.

Mr Davis 97 said:
I am not even sure how to show that ##\sin(\frac{n\pi}{3})## diverges...
The sine expression takes on only three different values: ##0, \frac{\sqrt 3} 2, -\frac{\sqrt 3} 2##. For n = 0, ..., 5, the sequence is ##0, \frac{\sqrt 3} 2, \frac{\sqrt 3} 2, 0, -\frac{\sqrt 3} 2, -\frac{\sqrt 3} 2##, and then repeats these terms endlessly . Can this sequence converge?

For the series, for ##r \in (\infty, 0]##, take a look again at what I said about the Nth term test for divergence.
 
  • #5
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Let's look at things with some numbers, say r = -1.
##\frac 1 {n^r} = \frac 1 {n^{-1}} = n##. What is ##\lim_{n \to \infty} \frac 1 {n^r}##?
What if r = -1/2? What if r = -2? Etc.

The sine expression takes on only three different values: ##0, \frac{\sqrt 3} 2, -\frac{\sqrt 3} 2##. For n = 0, ..., 5, the sequence is ##0, \frac{\sqrt 3} 2, \frac{\sqrt 3} 2, 0, -\frac{\sqrt 3} 2, -\frac{\sqrt 3} 2##, and then repeats these terms endlessly . Can this sequence converge?

For the series, for ##r \in (\infty, 0]##, take a look again at what I said about the Nth term test for divergence.
What I can see is that if ##r\in (-\infty, 0)##, then ##\lim_{n\to\infty}\frac{1}{n^r} = \infty##. I also see that ##\sin(\frac{n\pi}{3})## diverges because subsequneces can converge to different numbers. But I don't know how to combine these facts to show that the product diverges.
 
  • #6
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What I can see is that if ##r\in (-\infty, 0)##, then ##\lim_{n\to\infty}\frac{1}{n^r} = \infty##. I also see that ##\sin(\frac{n\pi}{3})## diverges because subsequneces can converge to different numbers. But I don't know how to combine these facts to show that the product diverges.
Can you show that for ##r\in (-\infty, 0)##, ##\lim_{n \to \infty} \sin(\frac{n\pi} 3)\frac 1 {n^r} \ne 0##? That's all you need to show per the Nth Term Test I cited.
 
  • #7
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Can you show that for ##r\in (-\infty, 0)##, ##\lim_{n \to \infty} \sin(\frac{n\pi} 3)\frac 1 {n^r} \ne 0##? That's all you need to show per the Nth Term Test I cited.
It's intuitively obvious. The sin oscillates while the other term blows up. So clearly it doesn't converge. But I'm not sure how to show this rigorously...
 
  • #8
LCKurtz
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It's intuitively obvious. The sin oscillates while the other term blows up. So clearly it doesn't converge. But I'm not sure how to show this rigorously...
Take a subsequence of ##\frac{k\pi} 3## so the sine doesn't oscillate and isn't zero.
 
  • #9
Ray Vickson
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What I can see is that if ##r\in (-\infty, 0)##, then ##\lim_{n\to\infty}\frac{1}{n^r} = \infty##. I also see that ##\sin(\frac{n\pi}{3})## diverges because subsequneces can converge to different numbers. But I don't know how to combine these facts to show that the product diverges.
There is a fundamental theorem which states that the infinite series ##\sum_n t_n## converges only if ##|t_n| \to 0## as ##n \to \infty.## So, terms going to zero is a necessary (but by no means sufficient) condition for convergence of the series. I would be shocked if your book or notes does not state and prove this very fundamental fact---it is the first order of business when discussing infinite series!

You can (and should) look at the values of ##\sin(\pi n/3)## for a few values of ##n = 1,2,3, \ldots##, to understand whether or not the presence of the sine factor really complicates things very much.
 
  • #10
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Take a subsequence of ##\frac{k\pi} 3## so the sine doesn't oscillate and isn't zero.
So I want to show that for ##r\in (-\infty, 0)##, the series diverges. If we look at the subsequence ##a_{6k-5} = \frac{\sqrt{3}}{2}\frac{1}{(6k-5)^r}##, we see that ##\lim_{k\to\infty}a_{6k-5}=+\infty##. Since a subsequence diverges to infinity, ##\lim_{n\to\infty}a_n \not = 0##, so the series diverges. Is this correct?
 
  • #11
LCKurtz
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So I want to show that for ##r\in (-\infty, 0)##, the series diverges. If we look at the subsequence ##a_{6k-5} = \frac{\sqrt{3}}{2}\frac{1}{(6k-5)^r}##, we see that ##\lim_{k\to\infty}a_{6k-5}=+\infty##. Since a subsequence diverges to infinity, ##\lim_{n\to\infty}a_n \not = 0##, so the series diverges. Is this correct?
That looks OK. But I wouldn't phrase it as ##\lim_{n\to\infty}a_n \ne 0## because that seems to suggest it has a limit but the limit isn't ##0##. It's better to just say ##\lim_{n\to\infty}a_n## does not exist to avoid the ambiguity.
 
  • #12
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That looks OK. But I wouldn't phrase it as ##\lim_{n\to\infty}a_n \ne 0## because that seems to suggest it has a limit but the limit isn't ##0##. It's better to just say ##\lim_{n\to\infty}a_n## does not exist to avoid the ambiguity.
I'm not really seeing how to do the ##0<r\le 1## case. Mark44 said one might use the comparison test, but I am not seeing what to compare it to...
 
  • #13
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I'm not really seeing how to do the ##0<r\le 1## case. Mark44 said one might use the comparison test, but I am not seeing what to compare it to...
I don't either. I thought that it might work by comparing it to a divergent p-series, but after some thought, I don't think that would work.

Here's a different direction to try. The ##\sin(\frac{n\pi}3)## factor repeats a cycle of length 6: ##0, \frac{\sqrt 3} 2, \frac{\sqrt 3} 2, 0, -\frac{\sqrt 3} 2, -\frac{\sqrt 3} 2##, so your series looks like ##\left(\frac{\sqrt 3} 2 (\frac 1 {1^r} + \frac 1 {2^r}) - \frac{\sqrt 3} 2(\frac 1 {4^r} + \frac 1 {5^r})\right) + \left(\frac{\sqrt 3} 2(\frac 1 {7^r} + \frac 1 {8^r}) - \frac{\sqrt 3} 2(\frac 1 {10^r} + \frac 1 {11^r})\right) + \dots##
##=\frac{\sqrt 3} 2 \left( (\frac 1 {1^r} + \frac 1 {2^r} - \frac 1 {4^r} - \frac 1 {5^r} ) + (\frac 1 {7^r} + \frac 1 {8^r} - \frac 1 {10^r} - \frac 1 {11^r}) + \dots \right)##. If you can figure out what the general term looks like, I don't think it would be hard to show that the series diverges for ##r \in [0, 1]##
 
  • #14
Ray Vickson
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I don't either. I thought that it might work by comparing it to a divergent p-series, but after some thought, I don't think that would work.

Here's a different direction to try. The ##\sin(\frac{n\pi}3)## factor repeats a cycle of length 6: ##0, \frac{\sqrt 3} 2, \frac{\sqrt 3} 2, 0, -\frac{\sqrt 3} 2, -\frac{\sqrt 3} 2##, so your series looks like ##\left(\frac{\sqrt 3} 2 (\frac 1 {1^r} + \frac 1 {2^r}) - \frac{\sqrt 3} 2(\frac 1 {4^r} + \frac 1 {5^r})\right) + \left(\frac{\sqrt 3} 2(\frac 1 {7^r} + \frac 1 {8^r}) - \frac{\sqrt 3} 2(\frac 1 {10^r} + \frac 1 {11^r})\right) + \dots##
##=\frac{\sqrt 3} 2 \left( (\frac 1 {1^r} + \frac 1 {2^r} - \frac 1 {4^r} - \frac 1 {5^r} ) + (\frac 1 {7^r} + \frac 1 {8^r} - \frac 1 {10^r} - \frac 1 {11^r}) + \dots \right)##. If you can figure out what the general term looks like, I don't think it would be hard to show that the series diverges for ##r \in [0, 1]##
The series actually converges when ##r = 1,## but showing that is a bit tricky.

In fact, the series can be written as
$$S = \frac{\sqrt{3}}{2} \sum_{k=0}^{\infty} \left( \frac{1}{(6k+1)^r}+\frac{1}{(6k+2)^r}-\frac{1}{(6k+4)^r} - \frac{1}{(6k+5)^r} \right) $$
 
  • #15
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Would Dirichlet's test work in this case? https://en.wikipedia.org/wiki/Dirichlet's_test

If we let ##a_n=\frac{1}{n^r}## and ##b_n=\sin\frac{n\pi}{3}##. Then we see that ##\frac{1}{(n+1)^r}\le\frac{1}{n^r}## and ##\lim_{n\to\infty}\frac{1}{n^r} = 0##. We also have that ##\sum_{k=1}^{n}\sin (\frac{n\pi}{3}) \le \sqrt{3}## for all ##n\in\mathbb{N}##.
 
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  • #16
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Would Dirichlet's test work in this case? https://en.wikipedia.org/wiki/Dirichlet's_test

If we let ##a_n=\frac{1}{n^r}## and ##b_n=\sin\frac{n\pi}{3}##. Then we see that ##\frac{1}{(n+1)^r}\le\frac{1}{n^r}## and ##\lim_{n\to\infty}\frac{1}{n^r} = 0##. We also have that ##\sum_{k=1}^{n}\sin (\frac{n\pi}{3}) \le \sqrt{3}## for all ##n\in\mathbb{N}##.
Dirichlet's Theorem requires that ##|\sum_{n = 1}^N b_n | \le M## for every positive integer N. With ##b_n = \sin(\frac{n\pi} 3)## that inequality isn't satisfied. Also, the theorem says that ##\sum_{n = 1}^\infty a_nb_n## converges, and you're trying to show that it diverges, at least for values of ##r \in (-\infty, 1)##.
 

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