# Σ_{n=1} ^{∞} {(4 – cos(n^2))/n^2} (Includes nicer-looking image.)

1. Nov 14, 2013

### s3a

1. The problem statement, all variables and given/known data
For the series Σ_{n=1} ^{∞} {(4 – cos(n^2))/n^2} (this series can also be seen by looking at TheSeries.png.),

which of the following is true?:

A. This series converges.

B. This series diverges.

C. The integral test can be used to determine convergence of this series.

D. The comparison test can be used to determine convergence of this series.

E. The limit comparison test can be used to determine convergence of this series.

F. The ratio test can be used to determine convergence of this series.

G. The alternating series test can be used to determine convergence of this series.

2. Relevant equations
N/A

3. The attempt at a solution
I think only A and D being true is the answer, but why isn't that the case?

I think that A is true, because the numerator of that series ranges from 3 to 5 both, and all the numbers in that interval are smaller than 1 and 1/n^2 is a convergent p-series.

I think that B is false, because A is true.

I think that C is false, because it involves a Fresnel integral which is material that is more advanced than that of this course, but Wolfram Alpha says that this does converge, so I'm not sure what to assume here. I'd appreciate some elaboration on this one.

I think that D is true, and it is the reason I used to justify A.

I think that E is false, because the value of the numerator ranges from 3 to 5, and since there is no one value, the limit doesn't exist.

I think that F is false, because the cosine functions make the limit unpredictable, therefore it does not exist.

I think that G is false, because the alternating series applies to functions where there is a predictable negative or positive sign, instead of a range of values which occasionally reach -1 or 1 (and other values in between).

Could someone please tell me why I'm wrong?

#### Attached Files:

• ###### TheSeries.png
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2. Nov 14, 2013

### Simon Bridge

$$S=\sum_{n=1}^\infty \frac{4-\cos n^2}{n^2}$$
I think you need to be more careful to check that the tests are valid - formally.
What is the definition of the test? Is there a specific list of conditions that make the test valid - then list them and check them off one at a time.

The series does converge.
So the the mistake lies in checking the validity of the tests (or the model answer).

Per your specific question: something leading to a process too advanced for you, does not mean that it is not a valid approach to the problem. It's just not one you'd be expected to use.
So - if the integral test may be used by someone, doesn't have to be you, to determine convergence of this series - then statement C is true.

3. Nov 14, 2013

### Office_Shredder

Staff Emeritus
4. Nov 14, 2013

### pasmith

In principle this test can be used, because all the terms of the series are non-negative. The difficulty is to find a suitable series to compare it to.

5. Nov 14, 2013

### s3a

So, does that mean that A, C and D are the only true statements?

About C, thanks to the Internet, I was "lucky" to have come across Fresnel integrals, but what if another student had seen this problem without seeing anything that wasn't covered in the course, how would he or she be able to answer this? Is there another way to determine the validity of statement C?

6. Nov 14, 2013

### s3a

Sorry, I saw the other messages later.

Office_Shredder, that was exactly what I was looking for! Thanks! :D

Pasmith, wouldn't the cosine “see to it” that the limit does not exist?

7. Nov 14, 2013

### s3a

So, wait, it IS only A and D that are true statements?

8. Nov 14, 2013

### haruspex

As pasmith wrote, the limit comparison test will work fine if you can find a suitable other series. E.g. consider a series which is merely double the given one. If you were to prove by some means that this new series converges then the limit comparison test could be used to show the given series converges. There could be a less trivial example. So I see no way to decide whether E can be used, short of successfully using it.

9. Nov 14, 2013

### s3a

Oh, so let's say, I could prove that Σ_{n=1} ^{∞} {(4 – cos(n^2))/(n-1)^2}, for example, converges by other means, then I could use the limit comparison test to show that Σ_{n=1} ^{∞} {(4 – cos(n^2))/n^2} converges, since the limit will be 1, right?

Okay, so it's A, D and E, then? (Sorry for asking a lot, I just want to make sure I understand what is being said well.)

10. Nov 14, 2013

### haruspex

Yes, but you do not know whether that is possible without actually doing it. So you cannot decide E (yet).

11. Nov 14, 2013

### Simon Bridge

Please note (others too) I made no claim about the truth or falsehood of statement C. I was pointing out the logical fallacy in the provided argument for it being false.

The argument did not work, that does not mean the statement C is true.

I suggested going through the definitions for each carefully and formally.
Making sure you understand what the tests do:
i.e. http://tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx
Suppose that f(x) is a continuous, positive and decreasing function on the interval ...​

Fortunately the others seem to have done, or are close to doing, that work for you.
Take it one step at a time.

Does it give you any ideas? ... You have to actually apply the test to show if it will work or not - nobody will tell you unless you try.

Last edited: Nov 14, 2013
12. Nov 14, 2013

### s3a

Are you saying that I am supposed to assume that I must be able to compare the series of the problem with another series whose convergence is known to me immediately (such as a regular p series instead of something like Σ_{n=1} ^{∞} {(4 – cos(n^2))/(n-1)^2}) for these kinds of problems? Knowing this would be very useful, not just for this problem, but for future ones as well.

Simon Bridge, I get the integral test situation now. About the limit comparison test, I did apply the test with my “made-up” Σ_{n=1} ^{∞} {(4 – cos(n^2))/(n-1)^2}, and I get 0 < c = 1 < ∞, but I'm assuming that I am restricted only to “simple series” like p series for problems like these … am I correct in thinking that that is that the case?

I know in “real life”, if I somehow know my “made-up” series converges, I could determine that the series of this problem converges as well, but coursework is not exactly “real-life”.

13. Nov 14, 2013

### Office_Shredder

Staff Emeritus
Can you elaborate why you think A and D can't be the correct answer? Is it an answer at the back of a textbook or something?

14. Nov 14, 2013

### haruspex

No. I'm saying that if you had a series bn that you knew by other means to be convergent and you could show an/bn converges then you could answer yes to E. But since you do not have such a series to hand that doesn't work. OTOH, there might be such a series bn that you could show to be convergent by other means - you just haven't thought of it yet. So neither can you say E is false.

15. Nov 14, 2013

### s3a

Sorry, I double-posted. (I'm really sorry about this, I keep only clicking once yet it keeps posting twice.)

16. Nov 14, 2013

### s3a

Office_Shredder, no, it's some exam-preparation problem for which we don't have the answer. Someone told me the answer was A and D, but I'm not sure if that's true. Is it?

Haruspex, I see, that makes sense. Having said that, wouldn't I need some numerator for my $b_n$ that cancels out the 4 – cos(n^2)? I could only think of a $b_n$ that has a 4 – cos(n^2) on its numerator to cancel out the 4 – cos(n^2) of $a_n$'s numerator in order for the limit to exist … so we could safely conclude that E is false, and that (like I've been told) only A and D are correct, right?

You guys seem more reliable, and I just want to make sure that I have the correct answer, because I don't want to assume I'm right for something where I'm not right, because I want to make sure I understand all the material.

17. Nov 14, 2013

### Simon Bridge

The series you pick for a comparison doesn't have to be simple - but you do have to know if it converges or diverges, and you'd normally have to be able to prove which it does.

The method is:
Put $a_n =$<your series>.
Cunningly select a $b_n$ where you know for sure what it does, and can prove it if anyone asks.
find the limit for $a_n/b_n$ ... if it is positive and greater than zero - the your series is the same type as the one you know.

The trick is to pick a $b_n$ so that it makes finding the limit simple.

It is kinda looking like the answer you have is correct isn't it.
But that's not the problem - the problem is, how would you be able to do this problem in an exam?
You need to be able to figure for yourself whether the answer is correct or not.
Next time it may not pop up in multi-choice.

18. Nov 14, 2013

### LCKurtz

Seems like a good idea. Have you understood Haruspex's post #8?

19. Nov 15, 2013

### s3a

I, at least, feel like I do. :)

I know, but these multiple-choice problems are good for me to look very hard at each test, instead of using any test to determine whether a series converges or not, and this problem has helped me do that!

20. Nov 15, 2013

### Staff: Mentor

You can use the integral test together with the comparison test to verify that the series over 5/n^2 and 3/n^2 converge.

Next https://www.physicsforums.com/forumdisplay.php?f=231 [Broken]: split the series into suitable subseries to use the alternating series test on the cos(n^2)-part :D.

Last edited by a moderator: May 6, 2017