# Homework Help: Comparison Test for Infinite Series Example

1. Sep 11, 2009

### philnow

1. The problem statement, all variables and given/known data
Using the comparison test determine if the infinite series for

sin(3/n^2)

converges or diverges.

3. The attempt at a solution

Well... these are pretty straight forward, and it's pretty obvious that this is convergent, but I'm having trouble applying the comparison test to a trig function. Can we say that:

sin(3/n^2) < sin(1/n)

which converges?

2. Sep 11, 2009

### Dick

i) Why do you think sin(3/n^2)<sin(1/n)? If n=1 that says sin(3)<sin(1). That's false. ii) Why do you think sin(1/n) converges?? That's also false. Give reasons why you think your statements are true! Hint: think about using sin(x)<x for x>=0.

3. Sep 11, 2009

### fmam3

It's actually pretty easy to see why the series $$\sum \sin( \frac{3}{n^2})$$ does not converge. Note that if $$\sum a_n$$ converges, then $$\lim a_n = 0$$. Can you see why $$\lim \sin(3 / n^2) = 0$$ is not true?

4. Sep 11, 2009

### Staff: Mentor

Unless there's something I'm not seeing,
$$\lim_{n \rightarrow \infty} sin(3/n^2)~=~0$$
but that isn't much help. The converse of the theorem you cited isn't true. IOW, there are many series for which an --> 0, but the series diverges. An example is the harmonic series, whose terms approach zero, even though the series diverges.

5. Sep 11, 2009

### philnow

In retrospect I don't know why I thought sin(1/n) converges. Still I can't come up with a good series to compare to the original...

6. Sep 11, 2009

### fmam3

That's exactly what I'm saying, isn't it? If $$\sum a_n$$ converges, then $$\lim a_n = 0$$.

For this case, for contradiction, suppose that the series $$\sum \sin(3 / n^2)$$ converges. Then we must have that $$\lim_{n \to \infty} \sin(3 / n^2) = 0$$. Then, $$\forall \varepsilon > 0, \exists N$$ such that $$n \geq N_1$$ implies $$| \sin(3 / n^2) - 0 | < \varepsilon$$. In particular, we would have $$| \sin(3 / n^2) | < 1/4$$ for $$n > N_1$$. However, we can pick $$n$$ such that $$\sin(3 / n^2) = \pi / 2 > 1/4$$ right? In particular, set $$3 / n^2 = \pi / 2 + 2\pi k$$ for some $$k \in \mathbb{Z}$$. Then from here, it should be quite straight forward to see that $$( \sin(3 / n^2) )_{n \in \mathbb{N}}$$ does not converge to zero and hence its associated series does not converge....

I guess an even slicker way of showing the above is to consider the limsup and liminf. It should be clear that $$\limsup \sin(3 / n^2)} = 1$$ and $$\liminf \sin(3 / n^2) = -1$$. And since $$\limsup \sin(3 / n^2} \ne \liminf \sin(3 / n^2)$$, it follows that the limit $$\lim \sin(3 / n^2)$$ does not exist, let alone equal to zero.

The harmonic series $$\sum 1/n$$ is not what I had claimed above. The harmonic series diverges, so the statement above does not even apply... (i.e. the hypothesis is that the series converges).

Last edited: Sep 11, 2009
7. Sep 11, 2009

WHY don't you think $$\sin(3/{n^2})$$ goes to zero? Also note that it is impossible for you to "pick n such that"

$$\sin(3/{n^2}) = \frac{\pi}{2}$$

since the fraction you give is larger than one.

You can't even do this: satisfy
$$3 / n^2 = \pi / 2 + 2\pi k$$

since the left side is rational while the right side is not. It is in fact true that

$$\lim_{n \to \infty} \sin\left(\frac{3}{n^2}\right) = 0$$

True, this does not mean you can conclude with certainty that the given series converges, but it does mean it is a possibility.

8. Sep 11, 2009

### Staff: Mentor

No, wrong. There is no real number n for which sin(3/(n2)) equals pi/2. None. It is very simple to show that sin(3/(n2)) converges to zero, but you're trying to convince me that this sequence does not converge, which is going to be difficult for you to do.

If you want me to give you a value of N for which n > N ==> |sin(3/(n2))| < 1/4, I can do that. Smaller than 1/10? I can do that too. Smaller than 0.01? I can do that. Smaller than 0.001? I can do that. And so on, as long as it takes to convince you that this sequence converges to zero.

9. Sep 11, 2009

Back to the original question about a comparsion test: do you know of a simple inequality that relates $$\sin(x)$$ and $$x$$ for $$x > 0$$? If so, that is the key to finding a suitable series to use in the comparison test.

10. Sep 11, 2009

### snipez90

Eh ok, let's not harp on fmam3 just because he used the "divergence test" or whatever it's called incorrectly. I mean yes the harmonic series is an obvious example of where the converse of the theorem in question fails, but usually it's mainly used to show divergence so you don't have to deal with other tests. In rare cases, you can in fact use it to show convergence of a series, but not directly.

Anyways, to the OP: if you didn't follow Dick's original hint, you could always use the limit comparison test (unless you are told to use only the regular comparison test) and it's actually pretty obvious what to "limit compare" the sequence of terms defining the original series to.

Last edited: Sep 11, 2009
11. Sep 11, 2009

"fmam3 just because he used the "divergence test" or whatever it's called incorrectly"

It's a little more serious than using a test incorrectly: the core of the argument, was that

$$\sin\left(\frac 3 {n^2}\right)$$

doesn't converge to zero, with few other errors thrown in. Making those statements in one post could be written off as minor, but defending them tends to be more serious. Does it reflect long-term? Probably not, but it was a serious case of mis-advice to this particular poster.

Last edited: Sep 11, 2009
12. Sep 11, 2009

### fmam3

Oh... sorry my bad to all of the guys above ---- I don't know what I was smoking when I thought $$\sin\left(\frac 3 {n^2}\right)$$ DOES NOT go to zero. I was a little caught up by the whole sinosoid and this fact $$|\sin\left(\frac 3 {n^2}\right)| \leq 1$$ thing.

But if there's any "merit" (or face-saving strategies for myself) out of this discussion, I have to just clarify this "divergence test" thing ---- it was actually not what I had tried to assert. I was merely trying to say, IF a series $$\sum a_n$$ converges, then $$\lim a_n = 0$$. But I'm not asserting the converse whatsoever. And furthermore, DOES NOT converge does not necessarily mean diverge anyway. For instance, $$a_n = (-1)^n$$ does not converge, but doesn't necessarily mean it diverges.

Truly, sorry for confusing the OP and other participants of this thread. My apologies.

13. Sep 11, 2009

It was clear to me you weren't trying to use the aforementioned test incorrectly - no problem there.

I'm not sure what you mean when you say $$(-1)^n$$ doesn't necessarily diverge - it does.

But no matter: thanks for your comment: it's behind all of us now. About "I don't know what I was smoking..." - I remember some times like that. (Alas, now, for me, Homer Simpson's classic lament "I used to rock and roll every night and party every day. Now I'm lucky if I get down once a month." holds true.) Sleep it off (attempt at humor here) and come back full strength tomorrow.

14. Sep 11, 2009

### fmam3

Thanks for understanding :P It's just one of those days.

Without trying to derail or hijack this thread any further, but I would like to respond to the point about $$(-1)^n$$. Perhaps its just entirely my fault or gaps in understanding (and if so, please do correct!), but I think there's a mis-communication between how we interpret "divergence". When I meant "divergence" there, I'd meant diverging to $$\pm \infty$$. So for this sequence $$(-1)^n$$, I should really say: it does not converge to a real number, and also does not diverge to $$\pm \infty$$.

I think I should just shut up now --- no more math for me today :P

15. Sep 11, 2009

### Dick

Wow, this isn't going well. philnow, are you ready to help us out, yet?

16. Sep 12, 2009

### Staff: Mentor

Divergence is the opposite of convergence. If a sequence diverges, all that means is that the sequence doesn't converge to a specific number. It can diverge by growing ever larger or ever more negative, or it can diverge by flip-flopping back and forth between two numbers, like your example of an = (-1)n.

17. Sep 12, 2009

### philnow

Wow, haha, what have I created?

Can I consider sin(x)<x for all x>=0 and then say that

sin(3/n^2) < 3/n^2

where 3/n^2 converges (power series 2)?

18. Sep 12, 2009

### snipez90

Yes that works, but since I tend to forget basic inequalities such as that one, I probably overuse limit comparison.

19. Sep 12, 2009

### philnow

Thanks for the confirmation, I would have used the LCT if the question permitted it >.<

20. Sep 12, 2009

### gregmoon

I am having a very similar problem, but for the general case of $$\sum \sin( \frac{x}{n^2})$$. I cannot think of a proper comparison for this. I'm not sure if the inequality stated above still holds.