# Finding the r value for the series to converge

## 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

## 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

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Mark44
Mentor

## 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

## 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.

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
Mentor
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.

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
Mentor
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.

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
Homework Helper
Gold Member
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
Homework Helper
Dearly Missed
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.

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
Homework Helper
Gold Member
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.

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

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 $n\in\mathbb{N}$.

Last edited:
Mark44
Mentor
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 $r \in (-\infty, 1)$.