# Limit of a sequence

1. Jan 31, 2006

### VietDao29

After browsing some maths forums, I come across some problems. But since that forum is quite unactive, so I think I'll post it here.
So this is not homework.
I have never seen this kind of problem before. I've tried all of them, and so I'm posting for some confirmation from you guys. There are 2 problems in total.
1. Prove the following limit does not exist:
$$\lim_{n \rightarrow + \infty} \sin (n)$$
Proof:
There exists infinity n's such that:
$$\frac{\pi}{6} + 2k \pi < n < \frac{5 \pi}{6} + 2k \pi, \ k \in \mathbb{Z}$$, it means:
$$\frac{1}{2} < \sin (n) < 1$$
There also exists infinity n's such that:
$$-\frac{5 \pi}{6} + 2k \pi < n < -\frac{\pi}{6} + 2k \pi, \ k \in \mathbb{Z}$$, it means:
$$- 1 < \sin (n) < - \frac{1}{2}$$
So as n tends to infinity, sin(n) can take either value from $$\left] -1 ; \ - \frac{1}{2} \right[$$, or $$\left] \frac{1}{2} ; \ \ 1 \right[$$. That yields the limit above does not exists.
-----------------
2. Prove the following limit does not exist:
$$\lim_{n \rightarrow + \infty} \sin (n ^ 2)$$
Proof:
If that limit exists then:
$$\lim_{n \rightarrow + \infty} \sin ((n + 1) ^ 2) - \sin (n ^ 2) = 0$$
That means:
$$\lim_{n \rightarrow + \infty} \sin ((n + 1) ^ 2) - \sin (n ^ 2) = 2 \lim_{n \rightarrow + \infty} \cos \left( n ^ 2 + n + \frac{1}{2} \right) \sin \left( n + \frac{1}{2} \right) = 0$$
Since $$\lim_{n \rightarrow + \infty} \sin \left( n + \frac{1}{2} \right)$$ does not exist (it can be proved exactly like number 1), it means that, if this limit: $$\lim_{n \rightarrow + \infty} \sin (n ^ 2)$$ exists then this limit: $$\lim_{n \rightarrow + \infty} \cos \left( n ^ 2 + n + \frac{1}{2} \right) = 0$$ must be true.
But if this limit $$\lim_{n \rightarrow + \infty} \cos \left( n ^ 2 + n + \frac{1}{2} \right) = 0$$ is true then: $$\lim_{n \rightarrow + \infty} \cos \left[ 2 \left( n ^ 2 + n + \frac{1}{2} \right) \right] = \lim_{n \rightarrow + \infty} \cos \left( 2 n ^ 2 + 2n + 1 \right) = \lim_{n \rightarrow + \infty} \cos ^ 2 \left( n ^ 2 + n + \frac{1}{2} \right) - \sin ^ 2 \left( n ^ 2 + n + \frac{1}{2} \right) = -1$$ must also be true.
But $\cos (\pi + 2k \pi) = -1, \ k \in \mathbb{Z}$
Define a sequence {nk}, such that ni = 2i2 + 2i + 1.
Also define a sequence {ux} : $$u_i \in \mathbb{Z ^ +}$$ such that:
$$\left| \pi + 2u_i \pi - n_i \right|$$ will return the smallest positive number possible.
That means, as i tends to infinity $$\left| \pi + 2u_i \pi - n_i \right|$$ must tend to 0.
Assume that's it's true, we have: we can choose a number $\epsilon > 0$, there will exist some N such that if p > N then:
$$\left| \pi + 2u_p \pi - n_p \right| = \left| \pi + 2u_p \pi - 2p ^ 2 - 2p - 1 \right| < \epsilon$$ If p > N, then of course p + 1 > N
$$\left| \pi + 2u_{p + 1} \pi - n_{p + 1} \right|$$
$$= \left| \pi + 2u_{p + 1} \pi - 2(p + 1) ^ 2 - 2(p + 1) - 1 \right| < \epsilon$$. But we have:
$$\left| \pi + 2u_{p + 1} \pi - 2(p + 1) ^ 2 - 2(p + 1) - 1 \right| = \left| \pi + 2u_{p + 1} \pi - 2p ^ 2 - 6p - 5 \right| = \left| (\pi + 2u_p \pi - 2p ^ 2 - 2p - 1) + 2u_{p + 1} \pi - 4p - 4 - 2u_p \pi \right| < \epsilon$$
Now if $\epsilon$ is very small $\pi + 2u_p \pi - 2p ^ 2 - 2p - 1$ will tend to 0. To keep the inequality true, $2u_{p + 1} \pi - 4p - 4 - 2u_p \pi = 2 \pi (u_{p + 1} - u_p) - 2p - 4[/tex] must also tend to 0. That means, as p tends to inifity: 2p + 4 will tend to some multiple of [itex]\pi$. But that's clearly wrong. So that means this limit:
$$\lim_{n \rightarrow + \infty} \cos \left( 2 n ^ 2 + 2n + 1 \right)$$
does not exist. So this limit:
$$\lim_{n \rightarrow + \infty} \sin (n ^ 2)$$ does not exist either.
Are my steps correct? Is there any other way?
Thanks... :)

Last edited: Jan 31, 2006
2. Jan 31, 2006

### VietDao29

I posted this thread at ysterday midnight, so I felt sleepy, and hence, I tried to finish the post as soon as possible.
After reading the post again, I see that there's something not very clear here. So I am going to prove this statement.
There will exist an even integer x = 2m, such that:
$$-3 + 2k \pi < 2m < -1 + 2k \pi$$.
This is true since $-3 + 2k \pi$ is an irrational number, and $(-1 + 2k \pi) - (-3 + 2k \pi) = 2$
Adding 4 to that inequality yields:
$$1 + 2k \pi < 2m + 4 < 3 + 2k \pi$$
This means there exists infinity p's, such that:
$$1 + 2k \pi < 2p + 4 < 3 + 2k \pi$$
That means:
$$\sin (1 + 2k \pi) > \sin (2p + 4) > \sin (3 + 2k \pi)$$
$$0.841 > \sin (2p + 4) > 0.141$$.
Do the same, and we can prove that there also exists infinity p's such that:
$$-3 + 2k \pi < 2p + 4 < -1 + 2k \pi$$
That means:
$$-0.141 > \sin (2p + 4) > -0.841$$
Again, from the two inequalities above, one can prove that:
$$\lim_{p \rightarrow \infty} \sin (2p + 4), \ p \in \mathbb{Z}$$ does not exist.
If (2p + 4) will tend to some multiple of $\pi$, that means $$\lim_{p \rightarrow \infty} \sin (2p + 4) = 0$$, which contradicts what we've shown earlier.
Thanks a lot,

3. Feb 1, 2006

### StatusX

This problem is more subtle than I thought. But I think I have found a shortcut to proving it. I posted something like this yesterday, but then I realized there was a hole in my argument. I think I've patched it up now.

You have proven that if the limit exists, we must have:

$$\lim_{n \rightarrow \infty} \cos(n^2+n+\frac{1}{2}) = 0$$

We also have:

$$\frac{1}{2} \left( \sin(n^2) + \sin( (n+1)^2) \right) = \sin (n^2+n+\frac{1}{2} ) \cos( n+\frac{1}{2} )$$

and if the limit exists, it must be equal to the limit of this expression. But since $\cos(n^2+n+\frac{1}{2})$ gets arbitrarily close to zero, $\sin(n^2+n+\frac{1}{2})$ must get arbitrarily close to either 1 or -1, ie, it may alternate between them. So we have only to show that $|\cos( n+\frac{1}{2} ) |$ does not converge.

Since there are infintely many n+1/2 between $k \pi - \frac{\pi}{6}$ and $k \pi + \frac{\pi}{6}$, there are infintely many such that $\frac{\sqrt{3}}{2} \leq |cos(n+\frac{1}{2})| \leq 1$. Similarly, since there are infinitely many n+1/2 between $k \pi + \frac{\pi}{3}$ and $k \pi + \frac{2\pi}{3}$, there are infintely many such that $0 \leq |cos(n+\frac{1}{2})| \leq \frac{1}{2}$. Thus |cos(n+1/2)| does not converge, and so neither does sin(n2).

Last edited: Feb 1, 2006
4. Feb 3, 2006

### StatusX

Sorry I didn't go through your whole proof before, it looked pretty daunting. But it seems to be valid. One suggestion I have is instead of switching to integer sequences and then going back to trig to show they can't exist, can you find a way to stay in trig functions the whole time?

Last edited: Feb 3, 2006
5. Feb 3, 2006

### VietDao29

That's okay, buddy
Your way seemed much shorter, and nicer than me . I dunno why I always complicate things... However, I believe I've found a flaw in my resoning:
Suppose, you have 2 sequences:
{xn}, and {yn} defined as below:
$$x_i := \left\{ \begin{array}{l} 1, \mbox{ if } i \mbox{ is even.} \\ 0, \mbox{ if } i \mbox{ is odd.} \end{array} \right.$$
$$y_i := \left\{ \begin{array}{l} 1, \mbox{ if } i \mbox{ is odd.} \\ 0, \mbox{ if } i \mbox{ is even.} \end{array} \right.$$
So you'll have:
$$\lim_{n \rightarrow \infty} x_n y_n = 0$$, and $$\lim_{n \rightarrow \infty} x_n$$ does not exist.
However, you'll also have $$\lim_{n \rightarrow \infty} y_n$$ does not exist, either, instead of $$\lim_{n \rightarrow \infty} y_n = 0$$.
-------------------
So this part of my proof is not quite correct:
I mean, there's a chance that this limit does not exist, either:
$$\lim_{n \rightarrow + \infty} \cos \left( n ^ 2 + n + \frac{1}{2} \right)$$
And I'm working on another way round to solve the problem that does not involve the fact that: $$\lim_{n \rightarrow + \infty} \cos \left( n ^ 2 + n + \frac{1}{2} \right) = 0$$ or I'm trying to prove that this fact is correct.
However I haven't found out the way ... So what do you think? Have you found another way?
Thanks, :)

Last edited: Feb 3, 2006
6. Feb 4, 2006

### StatusX

Hmm, I didn't notice that. Well, I've thought of a new approach that hopefully will work.

We'll assume that:

$$\lim_{n \rightarrow \infty} \sin (n^2) = \sin(\alpha)$$

for suitable choice of $\alpha$. Then:

$$\lim_{n \rightarrow \infty} \sin^2 (n^2) = \sin^2 (\alpha)$$

This is because of a general fact that, given a sequence an such that:

$$\lim_{n \rightarrow \infty} a_n = L$$

and a continuous function f(x), then:

$$\lim_{n \rightarrow \infty} f(a_n) = f(L)$$

This is obvious for the f(x)=x2 that we just used, but in a moment we'll apply this theorem to a more complicated function. Note that sin(mx) for integral m can in general be expressed as a polynomial in sin(x) and cos(x). In fact, for a given m, the power of sin(x) is either odd in every term or even in every term of this polynomial, and similarly for cos(x). Thus if we square this function, each power is multplied by a power of the same parity, and so every power of sin(x) and cos(x) in this new polynomial is even. Thus sin2(mx) it is a polynomial in sin2(x) and cos2(x), and so also just in sin2(x)=1-cos2(x). We define:

$$\sin^2(mx) = f_m (\sin^2(x) )$$

And so fm is a polynomial. Thus it is continuous, and so we get:

$$\lim_{n \rightarrow \infty} f_m( \sin^2 (n^2)) = \lim_{n \rightarrow \infty} \sin^2 (m n^2)= f_m( \sin^2 (\alpha) ) = \sin^2 (m \alpha)$$

But notice that if m is a perfect square, this is a subsequence of the sin2(n2) sequence, and so must have the same limit. So we get:

$$\sin^2 (\alpha) = \sin^2 (m^2 \alpha)$$

For all integers m. This can easily be rearranged to give:

$$\cos (2\alpha) = \cos (2m^2 \alpha)$$

To solve this, we use:

$$\cos(mx)-\cos(nx) = -2\sin(\frac{m-n}{2} x) \sin( \frac{m+n}{2} x )$$

So that cos(mx)=cos(nx) when this right side is zero, ie, at:

$$x=\frac{ 2k \pi}{m-n} \mbox{ or } x=\frac{ 2k \pi}{m+n}$$

for any integer k. It isn't hard to verify by checking a few m's (2 and 3 are pretty easy) that the only possible values for $2\alpha$ are $2 k\pi$, so that $\alpha$ is $k\pi$, and so the limit of the sin(n2) must be zero, if it exists.

Now:

$$\sin((n+1)^2) = \sin(n^2)\cos(2n+1) + \cos(n^2)\sin(2n+1)$$

This must have the same limit as sin(n2). If this limit is zero, the first term on the right vanishes, and cos(n2) goes to +/- 1. If the limit of sin(2n+1) isn't zero, we're done.

But for this to have a limit of zero, the difference between successive arguments of the sin, ie, (2(n+1)+1) - (2n+1) =2, must approach a multiple of $\pi$. It doesn't, and so sin(2n+1) doesn't have a limit of zero, and so there is no limit for sin(n2).

Last edited: Feb 4, 2006
7. Feb 4, 2006

### arildno

I haven't read through all the posts, but you make at one point a grave mistake:
Just because sin(n) has no limit, it does NOT follow that your cosine limit must be zero in order for the sin(n^2) limit to exist!

Let $a_{k}=(-1)^{k}, b_{k}=(-1)^{k}, c_{k}=a_{k}*b_{k}=1$
So, just because the product sequence has a limit and one of the factor sequences does not, it does not follow that the other factor sequence must tend towards zero.

EDIT:
Okay, I see now you are working with that.

Last edited: Feb 4, 2006
8. Feb 4, 2006

### VietDao29

Maybe I'm missing something, but from:
$$\sin^2 (\alpha) = \sin^2 (m^2 \alpha)$$
I can only rearrange it to give:
$$\cos (2\alpha) = \cos (2m^2 \alpha)$$

You forgot a factor 2 . It should read:
$$\sin(mx)-\sin(nx) = 2 \sin(\frac{m-n}{2} x) \cos( \frac{m+n}{2} x )$$

$$\cos(2(n+1)+1) - \cos(2n+1) = -2 \sin(2n + 2) \sin(1)$$.

What's (4(n+1)+3) - (4n+3) = 4?

Apart from that, I think everything looks good.
---------------------------------------
I think I found another simplier way to show that if the limit:
$$\lim_{n \rightarrow \infty} \sin (n ^ 2) = L$$ exists then it can only be 0.
Here's my reasoning:
We have:
$$\lim_{n \rightarrow \infty} \sin (n ^ 2) = \lim_{n \rightarrow \infty} \sin ((2n) ^ 2) = \lim_{n \rightarrow \infty} \sin (4n ^ 2) = L$$. (1)
Now let's expand sin(4n2), and cos(4n2) in terms of sin(n2), and cos(n2) (De'Moivre formula).
We have:
$$\sin (4n ^ 2) = 4 \cos ^ 3 (n ^ 2) \sin (n ^ 2) - 4 \cos (n ^ 2) \sin ^ 3 (n ^ 2)$$. (2)
$$\cos (4n ^ 2) = \cos ^ 4 (n ^ 2) - 4 \cos ^ 2 (n ^ 2) \sin ^ 2 (n ^ 2) + \sin ^ 4 (n ^ 2) = \cos ^ 4 (n ^ 2) - 4 \cos ^ 2 (n ^ 2) (1 - \cos ^ 2 (n ^ 2)) + (1 - \cos ^ 2 (n ^ 2)) ^ 2$$
$$= 6 \cos ^ 4 (n ^ 2) - 6 \cos ^ 2 (n ^ 2) + 1$$. (3)
From (1), and (2), we have:
$$L = 4 \cos ^ 3 (n ^ 2) L - 4 \cos (n ^ 2) L ^ 3$$ (4)
One solution to the equation (4) is L = 0.
If L is not 0, then divide both sides of (4) by L, and get:
$$1 = 4 \cos ^ 3 (n ^ 2) - 4 \cos (n ^ 2) L ^ 2 = 4 \cos ^ 3 (n ^ 2) - 4 \cos (n ^ 2) (1 - \cos ^ 2 (n ^ 2))$$
$$= 8 \cos ^ 3 (n ^ 2) - 4 \cos (n ^ 2)$$. Rearrange this yields the equation:
$$8 \cos ^ 3 (n ^ 2) - 4 \cos (n ^ 2) - 1 = 0$$ (5).
Equation (5) has 3 solutions, namely:
$$\left[ \begin{array}{l} \cos (n ^ 2) = -\frac{1}{2} \\ \cos (n ^ 2) = \frac{1 + \sqrt{5}}{4} \\ \cos (n ^ 2) = \frac{1 - \sqrt{5}}{4} \end{array} \right.$$.
Each solution will gives one different sin(n2). I means there are no solution that are opposite number of each other.
That means, if $$\lim_{n \rightarrow \infty} \sin (n ^ 2)$$ does exist , and not 0, then cos(n2) does converge to 1 of the three solutions above.
Plug the three solutions above to (3), and use the fact that:
$$\lim_{n \rightarrow \infty} \cos (n ^ 2) = \lim_{n \rightarrow \infty} \cos (4n ^ 2)$$ (6).
None of the 3 solutions above satisfied the condition (6). Hence, if sin(n2) does converge to some value, it must then be 0.
From there, we can use the last part of your argument to show that $$\lim_{n \rightarrow \infty} \sin (2n + 1) \neq 0$$, and finish the proof. (Q.E.D)
But I don't really understand the proof that says: $$\lim_{n \rightarrow \infty} \sin (2n + 1) \neq 0$$ (I have asked some question above) can you clarify it?
Or I think we can also use the method in post #2 to show that: $$\lim_{n \rightarrow \infty} \sin (2n + 1) \neq 0$$, too.
Does this look correct? Any flaws found?
Thanks, :)

Last edited: Feb 4, 2006
9. Feb 4, 2006

### VietDao29

Yes, I've realised that, and I'm trying hard to correct that grave mistake.
It may be probably the result of solving problems at midnight. :)
Thanks, :)

Last edited: Feb 4, 2006
10. Feb 4, 2006

### StatusX

Sorry, that should be (2(n+1)+2)-(2n+2)=2. The 4n+3 is left over from a slightly different way I had done it before (showing that the absolute value has no limit, which turns out not to be necessary).

Yea, sorry about that. It was late and I wasn't too hung up on the details (eg, "or something like that"). But those don't seem to significantly alter the proof. I'll get around to fixing them a little later.

I haven't checked all your algebra, but one thing I noticed is:

$$\cos (4n ^ 2) = \cos ^ 4 (n ^ 2) - 4 \cos ^ 2 (n ^ 2) \sin ^ 2 (n ^ 2) + \sin ^ 4 (n ^ 2)$$

should be:

$$\cos (4n ^ 2) = \cos ^ 4 (n ^ 2) - 6 \cos ^ 2 (n ^ 2) \sin ^ 2 (n ^ 2) + \sin ^ 4 (n ^ 2)$$

Also, you seem to be doing a lot of messy work with numbers where a more abstract argument would be neater. Just a suggestion.

EDIT: Ok, I fixed the proof. It actually turns out to be a little simpler this way.

Last edited: Feb 4, 2006
11. Feb 5, 2006

### VietDao29

I don't really understand this part. I get:
$$2 \alpha = \frac{2k \pi}{1 \pm m ^ 2}$$
m = 2, we have:
$$2 \alpha = \frac{2k \pi}{1 + 4} = \frac{2k \pi}{5}$$
or
$$2 \alpha = \frac{2k \pi}{1 - 4} = -\frac{2k \pi}{3}$$
m = 3, we have:
$$2 \alpha = \frac{2k \pi}{1 + 9} = \frac{k \pi}{5}$$
or
$$2 \alpha = \frac{2k \pi}{1 - 9} = -\frac{k \pi}{4}$$
It's not a multiple of $$\pi$$.
May be I am missing something, ain't I?
Thanks, :)

12. Feb 5, 2006

### VietDao29

Whoops, why can't I edit my previous post?
-----------------------
Okay, I'll post my correction to my damn long approach, hope someone may give it a glance and point out my errors.
I'm going to prove that $$\lim_{n \rightarrow \infty} \sin (n ^ 2) = L$$ exists then it can only be 0.
We have:
$$\lim_{n \rightarrow \infty} \sin (n ^ 2) = \lim_{n \rightarrow \infty} \sin ((2n) ^ 2) = \lim_{n \rightarrow \infty} \sin (4n ^ 2) = L$$. (1)

Now let's expand sin(4n2), and cos(4n2) in terms of sin(n2), and cos(n2) (De Moivre formula).
We have:

$$\sin (4n ^ 2) = 4 \cos ^ 3 (n ^ 2) \sin (n ^ 2) - 4 \cos (n ^ 2) \sin ^ 3 (n ^ 2)$$. (2)

$$\cos (4n ^ 2) = \cos ^ 4 (n ^ 2) - 6 \cos ^ 2 (n ^ 2) \sin ^ 2 (n ^ 2) + \sin ^ 4 (n ^ 2) = \cos ^ 4 (n ^ 2) - 6 \cos ^ 2 (n ^ 2) (1 - \cos ^ 2 (n ^ 2)) + (1 - \cos ^ 2 (n ^ 2)) ^ 2$$
$$= 8 \cos ^ 4 (n ^ 2) - 8 \cos ^ 2 (n ^ 2) + 1$$. (3)

-------------------

From (1), and (2), we have:
$$L = 4 \cos ^ 3 (n ^ 2) L - 4 \cos (n ^ 2) L ^ 3$$ (4)

One solution to the equation (4) is L = 0.

If L is not 0, then divide both sides of (4) by L, and get:
$$1 = 4 \cos ^ 3 (n ^ 2) - 4 \cos (n ^ 2) L ^ 2 = 4 \cos ^ 3 (n ^ 2) - 4 \cos (n ^ 2) (1 - \cos ^ 2 (n ^ 2))$$
$$= 8 \cos ^ 3 (n ^ 2) - 4 \cos (n ^ 2)$$. Rearrange this yields the equation:
$$8 \cos ^ 3 (n ^ 2) - 4 \cos (n ^ 2) - 1 = 0$$ (5).

Equation (5) has 3 solutions, namely:
$$\left[ \begin{array}{l} \cos (n ^ 2) = -\frac{1}{2} \\ \cos (n ^ 2) = \frac{1 + \sqrt{5}}{4} \\ \cos (n ^ 2) = \frac{1 - \sqrt{5}}{4} \end{array} \right.$$.

Each solution will give one different sin(n2). I mean none of the 3 solutions above are opposite number of each other.

That means, if $$\lim_{n \rightarrow \infty} \sin (n ^ 2)$$ does exist, and not 0, then cos(n2) must converge to 1 of the three solutions above.
Plug the three solutions above to (3), and use the fact that:
$$\lim_{n \rightarrow \infty} \cos (n ^ 2) = \lim_{n \rightarrow \infty} \cos (4n ^ 2)$$ (6).

Only 1 of the 3 solutions above satisfied the condition (6).
That solution is:
$$\cos (n ^ 2) = -\frac{1}{2}$$.
Now if sin(n2) converge, then it can only converge to one of the two following values:
$$\lim_{n \rightarrow \infty} \sin (n ^ 2) = \frac{\sqrt 3}{2}$$
Or
$$\lim_{n \rightarrow \infty} \sin (n ^ 2) = -\frac{\sqrt 3}{2}$$

First case:
$$\lim_{n \rightarrow \infty} \sin (n ^ 2) = \frac{\sqrt 3}{2}$$
and
$$\lim_{n \rightarrow \infty} \cos (n ^ 2) = -\frac{1}{2}$$
If those 2 limits really exists, then the difference between 2 successive arguments for the sin, and cos must be some multiple of $$2 \pi$$.
Now subtract them, we have: (n + 1)2 - n 2 = 2n + 1.
If as n tends to infinity, 2n + 1 tends to some multiple of $$2 \pi$$, then n + 1 / 2 will tend to some multiple of $$\pi$$, which is false because:
$$\lim_{n \rightarrow \infty} \sin \left( n + \frac{1}{2} \right)$$ does not exist (This can be proved just like the proof for Problem 1).

Second case:
$$\lim_{n \rightarrow \infty} \sin (n ^ 2) = -\frac{\sqrt 3}{2}$$
and
$$\lim_{n \rightarrow \infty} \cos (n ^ 2) = -\frac{1}{2}$$

Can be shown to be invalid exactly as the first case.

Hence, if sin(n2) does converge to some value, it must then be 0.
From there, we can use the last part of your argument to show that $$\lim_{n \rightarrow \infty} \sin (2n + 1) \neq 0$$, and finish the proof. (Q.E.D)
Can anyone fid any more flaws in my reasoning?
Thanks, :)

Last edited: Feb 5, 2006