An example of a very strange sequence

**Thommas** · August 11th 2011, 07:08 AM

Hallo everyone!

I hope, you can help me with the following problem:

Given the sequence (sin(n) )^n

Does is converge?

I already know, that the sequence sin(n)
(without the exponent)
is dense in [-1,1], since pi is irrational.

And I know that there is a subsequence of (sin(n))^n that converges to zero, meaning that
IF there is a limit, it has to be zero.

But so far, I have no clue, why there should be a limit. My guess is that there is no limit and that
(sin(n))^n
is also dense in [-1,1], but I have no idea if thats the case and how to prove it...

I hope you can help me!

Greetings,
Thomas :-)

**skeeter** · August 11th 2011, 09:35 AM

for $n \in \{0,1,2,3,...\}$

$|\sin(n)| < 1$ , since $n$ is never an odd multiple of $\frac{\pi}{2}$

so it would seem to me that the sequence $a_n = \{\sin^n(n) \}$ would converge to 0

**Thommas** · August 11th 2011, 09:52 AM

Thank you for your quick response.

However, I have a little problem with your argument:

For example the sequence
1-1/n
is also strictly less than 1.

So, following your argument, (1-1/n)^n would converge to 0.

But it converges to 1/e...

The problem is that sin(n) is never equal to 1 but comes very close to 1.

Greetings,
Thomas

**skeeter** · August 11th 2011, 10:54 AM

Originally Posted by Thommas

Thank you for your quick response.

However, I have a little problem with your argument:

For example the sequence
1-1/n
is also strictly less than 1.

So, following your argument, (1-1/n)^n would converge to 0.

But it converges to 1/e...

The problem is that sin(n) is never equal to 1 but comes very close to 1.

maybe that is the key to showing the sequence diverges.

it's been so long since I've studied analysis ... I'm sure one of the more learned individuals on the board will provide us both w/ some insight.

**Opalg** · August 11th 2011, 12:38 PM

Originally Posted by Thommas

Given the sequence (sin(n) )^n

Does it converge?

...

My guess is that there is no limit and that (sin(n))^n is also dense in [-1,1], but I have no idea if thats the case and how to prove it...

I think that you are right, and that this sequence does not converge to 0.

First idea: $\bigl(1-\tfrac1n\bigr)^n\approx e^{-1}$ . So if you can find a sequence of integers $(n_k)$ with $\sin(n_k)>1-\tfrac1{n_k}$ , then $(\sin(n_k))^{n_k}\approx e^{-1}$ , and so the subsequence $((\sin(n_k))^{n_k})$ will not converge to 0.

Second idea: To get $\sin(n_k)>1-\tfrac1{n_k}$ , you will need $n_k$ to be very close to an integer multiple of $\tfrac{\pi}2$ of the form $(4m+1)\tfrac{\pi}2.$ Equivalently, $\pi$ must be very close to a rational number of the form $\tfrac{2n_k}{4m+1}.$

That suggests looking at some known expressions for rational approximations to $\pi$ , such as the arctan formula, where you can control the accuracy of the approximation in terms of the numerator or denominator of the rational approximation.

I haven't tried to see whether these ideas can be made into a rigorous argument, but I think it's worth exploring.

**SammyS** · August 11th 2011, 05:02 PM

Originally Posted by skeeter

for $n \in \{0,1,2,3,...\}$

$|\sin(n)| < 1$ , since $n$ is never an odd multiple of $\frac{\pi}{2}$

so it would seem to me that the sequence $a_n = \{\sin^n(n) \}$ would converge to 0

I thought this sounded pretty good initially, but since sin(n)is dense in [-1,1], as n gets large, occasionally there will be an integer, k, close enough to being an odd multiple of π/2 so that sin(k) is very close to ±1; close enough so that even (sin(k))^k is very close to ±1 .

Some links to plots from WolframAlpha:

n from 1 to 60

n from 101 to 300

n from 1501 to 2500

**Melsi** · August 14th 2011, 02:11 PM

I too believe that sin(n) does not converge at all, although it is a bounded sequence. This is because sin is perodical function, and keeps coming round and round for ever as n goes to infinity.

And a last thing, all subsequences must converge to the same number so the sequence converges too.

**Thommas** · August 17th 2011, 08:25 AM

Thanks so much for the many ideas!

In particular, thanks Opalg for your two ideas.

In seems that in order to follow the first idea, one has to follow the second idea, which means one has to consider rational approximations of pi.

I took - as you suggested - the arctan-formula and did some calculations, but got into nowhere...

Still, this seems like a possibility.

Anyway, thank you all,
Bye,
Thommas

**Melsi** · August 17th 2011, 02:00 PM

$(1) \; b_{n}=sin \; floor(2n\pi + \pi/2)$ is a valid subsequence of $sin(n)$ containg an infinite number of terms.

$(2) \; c_{n}=sin \; ceil(2n\pi - \pi/2)$ is a valid subsequence too.

$(3) \; [floor(2n\pi + \pi/2) \geq 2n\pi + \pi/2 -1) ] \Rightarrow [sin \;floor(2n\pi + \pi/2) \geq sin (2n\pi + \pi/2 -1)=0.540302306 ]$

$(4) \; [ceil(2n\pi - \pi/2) \leq 2n\pi - \pi/2 +1) ] \Rightarrow [sin \;ceil(2n\pi - \pi/2) \leq sin (2n\pi - \pi/2 +1)=-0.540302306 ]$

$(5)$ An infinite number of terms are greater or equal than $sin(2n\pi + \pi/2 -1)=0.540302306$ , because of $(1)$ and $(3)$ .

$(6)$ An infinite number of terms are less or equal than $sin(2n\pi - \pi/2 +1)=-0.540302306$ , this is because of $(2)$ and $(4)$ .

$(7)$ Because of $(5)$ and $(6)$ an infinite number of terms cannot aproach each other as close as possible which means that the sequence $sin (n)$ cannot be a Cauchy Sequence!

The 1st Criterion of Cauchy says:
Any convergent sequences is a Cauchy Sequence. Because of $(7)$ this criterion is not satisfied.

This make sin(n) a sequence that has not limit at all in R or in +,- infinity.

I ask for understanding if I am wrong, just believed that I could help!

**Opalg** · August 18th 2011, 02:54 AM

Originally Posted by Thommas

In seems that in order to follow the first idea, one has to follow the second idea, which means one has to consider rational approximations of pi.

I took - as you suggested - the arctan-formula and did some calculations, but got into nowhere...

Still, this seems like a possibility.

I agree, the arctan formula is not good enough. But I think that it should be possible to prove the result using continued fractions.

Start with the Taylor expansion for $\cos x$ , which shows that $\cos x\geqslant 1-\tfrac12x^2.$ In particular, if $x <\sqrt{2/n}$ then $\cos x \geqslant 1-\tfrac1n.$ The behaviour of the sine function near a point of the form $\bigl(2k+\tfrac12\bigr)\pi$ is the same as the behaviour of the cosine function near 0. Therefore $\sin n > 1-\tfrac1n$ whenever $\bigl|n- \bigl(2k+\tfrac12\bigr)\pi\bigr| <\sqrt{2/n},$ or equivalently whenever $\Bigl|\pi - \frac{2n}{4k+1}\Bigr| < \frac{4}{(4k+1)\sqrt {2n}}.$

If we put $p=2n$ and $q=4k+1$ then $p/q$ is close to $\pi$ , and so p and q have the same order of magnitude. Up to some constant, therefore, we want to find values of p and q such that $|\pi - \tfrac pq| < 1/q^{3/2}.$ But the theory of continued fractions says that if p/q is a convergent for the continued fraction of $\pi$ , then $|\pi - \tfrac pq| < 1/q^2.$ That leaves plenty of room to spare, since $1/q^2$ will be much smaller than $1/q^{3/2}.$ Thus there are infinitely many values of p and q satisfying our requirements.

There is one niggling snag with that argument. Since $p=2n$ and $q=4k+1$ , we want to know that there are infinitely many convergents in the continued fraction expansion of $\pi$ in which the numerator is even and the denominator is a multiple of 4 plus 1. I am sure that this must be the case, but I have no idea how to prove it. The continued fraction expansion of $\pi$ is very erratic and unpredictable, and there seems to be not much known about it.

Footnote. It says here that $\frac{428224593349304}{136308121570117} =3.14159265358979323846264338327(569...)$ (correct to 29 digits). According to the above analysis, $(\sin n)^n$ should be well away from 0 when n is half the numerator of that fraction, namely $n=214112296674652.$

**Thommas** · August 18th 2011, 07:43 AM

Thanks Melsi, Thanks Opalg!:-)

First to Melsi:
Thanks for the beautiful and colorful picture. But I think your argument just shows that sin(n) does not converge. But I already know, that sin(n) is dense in [-1,1].
The original sequence was (sin(n))^n. So, your sequence to the power of n. And here the situation is more complicated.

which gets me to you, Opalg!
Thanks for your continued fractions. I will have to look at the argument more closely, since I'm not at all an expert on continued fractions.

Greetings,
Thomas

**Melsi** · August 18th 2011, 09:49 AM

I think my initial thought which said: if sin n does not converge then (sin n) ^n does not too might be wrong.
However I guess one can use the proof for sin(n).

If you are able to proove this hypothesis:

$(1) \; (\pi/2)^{n}\notin \mathbb{N}$ [HYPOTHESIS] then the following are true...

$(2) \; sin(n) \neq sin((\pi/2)^n)=\pm 1$ Because of $(1)$ .

$(3) \; -1<sin(n)^{n}<1$ because of $(2)$ .

$(4) \; a^{n}\rightarrow 0 \; because \; 0<a<1$ Because of Real number properties If remember well!

$(sin \; n)^{n} \to 0$ because of $(4) \; and \; (3).$

So if you are able to proove the ypothesis in red then your sequence $(sin \;n)^{n}$ should converge to zero as $n$ goes to infinity.

If you proove the oposite of the hypothesis then it diverges, I tested on pc and indications go for converging to zero, but you can never be sure using only PC!

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It is enough to proove that sin(n) does not converge.

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