Finding the r value for the series to converge

[Mr Davis 97]

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?

 

[Mark44]

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.

 

[Mr Davis 97]

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...

 

[Mark44]

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.

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.

 

[Mr Davis 97]

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.

 

[Mark44]

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.

 

[Mr Davis 97]

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...

 

[LCKurtz]

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.

 

[Ray Vickson]

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.

 

[Mr Davis 97]

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?

 

[LCKurtz]

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.

 

[Mr Davis 97]

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...

 

[Mark44]

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 $r \in [0, 1]$

 

[Ray Vickson]

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
$$S = \frac{\sqrt{3}}{2} \sum_{k=0}^{\infty} \left( \frac{1}{(6k+1)^r}+\frac{1}{(6k+2)^r}-\frac{1}{(6k+4)^r} - \frac{1}{(6k+5)^r} \right) $$

 

[Mr Davis 97]

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 $n\in\mathbb{N}$.

 

[Mark44]

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 $r \in (-\infty, 1)$.