Dear contributors, I have searched on internet, and found .... not our sequences, but their inverses. Those thoughts can move us forward:

https://www.quora.com/How-do-I-show-that-the-sequence-1-n-sin-n-diverges

https://math.stackexchange.com/questions/665776/convergence-of-the-sequence-frac1n-sinn

If I am not mistaken, this paper has not been published: https://arxiv.org/pdf/1104.5100.pdf

And yet! https://www.quora.com/Does-the-sequence-n-sin-n-have-a-convergent-subsequence

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

EDIT

I deleted my first answer because

Because of 1. and 2. it became misleading information.

I am attaching a pdf file with the proof that the finite limits are out of the question.

In this specific case, there is no convergence either with or without absolute values. However, generally when the series converges, removing the absolute value leads to a faster convergence, because negative and positive items in the sum partially compensate each other. A good example is the harmonic series - it does not converge: 1 + 1/2 + 1/3 + 1/4 + .... = Infinity

A similar series with alternating signs (the alternating harmonic series) converges: 1 - 1/2 + 1/3 - 1/4 + ... = Ln 2 (natural logarithm of 2)

This series converges, but there is no absolute convergence

I am also agree with Igor Ravve too. However., for practical applications via series approximation, we ignore the complex part and recourse to consider the real part for approximate solution. The complex or shadow part of the function affects the uniqueness of the solution (one or single solution) or convergence, exact solution, analytic solution or closed form solution (almost similar concepts).

As those sequences have no finite limits (see my proof above, looks correct), I think we need another proof to decide between "no limit" or "infinite limit".

Dear Viera,

I think, like you do, that we need a good solid argument for this problem. I tried the usual suspects (the epsilon-rank and epsilon-delta methods, subsequences, inequalities, trigonometric and density tricks), nothing (that I know of) worked.

I am amazed that I did not know of this problem until now.

All the best and Keep in touch,

George

Limits of sequences are closely related to limits of functions of one variable when x is tending to plus infinity. If the limit of f(x) is either finite or infinite, the limit of f(n) is the same (natural numbers are one of infinitely many sequences of x_n which converges to the unique limit). But it doesn't need to be the same in case of "multiple limits". For example f(x)=x-ent(x) (the difference between a real number and its integer part) converges to the interval [0,1) but the sequence f(n)=n-ent(n)=0 is constant and obviously has the limit 0.

The functions sin x and cos x obviously have neither finite nor infite limit at plus infinity - they converge to the interval [-1,1] (and |sin x| and |cos x| to [0,1]).

When multiplying by x, x·sin x and x·cos x converge to the interval (-∞,+∞) (and x·|sin x| and x·|cos x| to [0,+∞)).

So, we know in what intervals we can expect possible limits of sequences under consideration. But do any limits exist? It would mean that we can split natural numbers into subsequences nk of natural numbers for which subsequences eg. sin(nk) converge to some specific numbers.

You are correct, Przemysław, the function f(x)= x·sin x has no limit as x approaches +∞. Indeed, consider the sequences 2πn and, respectively π/2 + 2πn; we have that f(2πn)=0, which converges to 0, whereas f(π/2 + 2πn)=π/2 + 2πn, which converges to +∞.

If we assume that {n cosn} is convergent with limited L, then cosn=(1/n)(n cosn) tend to 0. But {cosn} is density in [-1, 1]. So this way is recived conventional.

If we use the fact that every convergent sequence is bounded and the converse is false, then we could finalize that the sequence is divergent.

I understand, Ayobami, yet I would like to know if "divergent" means "no limit" or "infinite limit".

Thus, Mehdi Mesrizadeh and Ayobami Salau have closed a proof by contradiction that if for some a>0 the limit of n^a sin(n) (etc.) exists, then it equals either oo or -oo. Good job! (after correction at 2020-07-06-23:37 in PL)

continued:

By the property that the limit points form the interval [-1;+1] one can infere that for arbitrary a>0, neither n^a sin(n) nor n^a cos(n) possesses the unbounded limit (be it +oo or -oo). Indeed, choosing n_k->oo and m_k->oo as k->oo in such a way that eg. sin(n_k) -> 1 and sin(m_k) -> -1 we arrive at the possibility, that

(n_k)^a sin(n_k) -> +oo and (m_k)^a sin(m_k) -> -oo

If the above is correct, then only the absolute values remain to be solved.

I am trying to explain reason which might be responsible for what Viera Čerňanová has reminded:

>> Nevertheless, if I remember well, na|sin(n)| goes to infinity for "a" high enough. Therefore, I am not sure if my argument above is quite strong.

You are correct Joachim, the absolute value case needs further attention.

Joachim Domsta "for obtaining the limit +oo for n^a |sin(n)| as n->oo, we need to go with the close to 0 values of sin(n) by multiplication by n^a sufficiently far away from zero. Yes, but how find this minimal a?"

According to a 1837 theorem by Gustav Dirichlet (about approximating general irrational numbers by continued fractions), the sequence

{ n1/2 sin(n) }

will contain an infinite subsequence which stays finite: Consider the sequence {pj/qj} of continued fraction approximations to π. The error of each approximation is less than 1/qj2, according to the Dirichlet theorem. This means that |sin(pj qj)| < 1/qj, which implies that sqrt(pj qj) |sin(pj qj)| < 2. Hence a ≥ 1/2. For sharper results one needs specific properties of rational approximations to π. The full continued fraction approximation is not known.

Correction: As becomes clear from the links posted by Viera Čerňanová, I do not make proper use of the Dirichlet theorem above, which says that pj = qj π + O(1/qj), which in turns implies that there is an infinite bounded subsequence {pj | sin pj |} of {n | sin n |}. It follows that none of the sequences in the question has a (finite or infinite) limit.

Thank you, Kåre! Any idea what can we say for n·|sin(n)| ?

My guess is that for arbitrary a>1 n^a |sin(n)| -> oo as n->oo.

It is based on the property that for any irrational positive number r the fractions of its multiplies frac( n r ) behave (with respect to some features) like a sequence of iid (independent identically distributed) random variables U_n with uniform distribution on the unit interval [0;1]. More precisely, the conjecture is true if frac(n/\pi) behaves like a typical probe of the sequence U_n. If this were true, then we can use that

(*) Pr{ n^a U_n -> oo } =1 for any a>1.

Due to trouble with my editors I am omitting any sketch of (*). By the way, it is based on the Borel-Cantelli theorem and therefore somehow available from public sources. When I get rid of my trouble and if still is needed, I will try to present a standard proof of (*).

Thank you Joachim; your argument might even work for a ≥ 1/2, like in Kåre's approach, maybe using the central limit theorem instead of Borel-Cantelli.

Thanks you George for kind suggestion, but Kåre's argument proves existence of bounded subsequence which would contradict existence of the unbounded limit, of course:-)

Best, Joachim

Because { cosn } is density on [-1,1], {n^a cosn} is density for every a>0 on over real line.

To Mehdi Mesrizadeh

It is not that simple. I think, the following might be a counterexample,

i.e. an example of a sequence with a dense set of values in [0;1], with non-dense values after multiplication by n:

y_n := exp_2{ - frac[ log_2(n) ] }, n=1,2,3, ...

where exp_2(z):= 2^z = 2z, log_2(w) = log2(w), w >0 (the inverse of exp_2 ).

Looking to the first sequence under attention, we see that it has a subsequence "converging" to infinity. To see this, observe that on each interval [2k*pi +pi/4, 2k*pi + 3*pi/4] (whose length is pi/2>1), there exists an integer n(k) at least. On the same interval, the values of the "sin" function are >= (1/2)**(1/2)=const>0. Consequently, the product converges to infinity, when k goes to infinity. The other sequences can be treated in a similar way.

Thank you, Octav! If we could find another sequence, with each term in an interval of length >1, that converges to anything but +∞, will do the job (conclusion: no limit). Not explicitly, but with a Dirichlet-type argument, as Kåre argued before.

Dear George, for the sequence n*sin(n) it is easy to find a subsequence which has exact the opposite values to those of the subsequence from above, so that it converges to - infinity. In case of n*|sin(n)| I do not have an answer in this respect for now.

George Stoica "Any idea what can we say for n·|sin(n)| ?"

It appears that nothing rigorous can be said about that! The question is really whether this sequence has an infinite subsequence of bounded values; otherwise the sequence can be said to "converge" to infinity.

This depends on "how irrational π is". Let x be an irrational number, and C = {pj/qj} its infinite sequence of continued fraction approximates, and let μ be the "largest" real number such that C has an infinite subsequence

S = {pα/qα } such that |x - pα/qα| ≤ 1/qαμ .

μ is essentially (the exact definition is somewhat different) the irrationality measure for x. Now, if π has irrationality measure μ, then there exist an infinite subsequence

{ nα = pαqα } such that nαa |sin(nα)| remains bounded for all a ≤ (μ-1)/2.

Hence, it becomes a question of the irrationality measure for π. By the Dirichlet theorem μ ≥ 2. According to a recent paper by Doron Zeilberger and Wadim Zudilin the best upper bound is Article The irrationality measure of π is at most 7.103205334137...

Comments: From what I have read today, almost all (apart from a set of measure zero) irrational numbers have irrationality measure equal to 2. So, probabilistically this is very likely to be the case for π also.

This belief is not changed by the known behaviour of the first 15 000 000 000 terms of the continued fraction approximations to π (computed by E.W. Weisstein) https://mathworld.wolfram.com/PiContinuedFraction.html.

In certain sense, it may very well be that 355/113 is the "absolutely best" rational approximation to π, https://www.quora.com/What-is-the-best-rational-approximation-of-pi-Let-best-be-the-difference-between-the-number-of-digits-used-to-represent-the-rational-and-the-number-of-accurate-digits-in-the-decimal-expansion

However, none of these considerations, or similar ones, constitutes a mathematical proof.

Correction: As becomes clear from the links posted by Viera Čerňanová , I do not make proper use of the Dirichlet theorem above, which says that pj = qj π + O(1/qj), which in turns implies that there is an infinite bounded subsequence {pj | sin pj |} of {n | sin n |}. It follows that none of the sequences in the question has a (finite or infinite) limit.

Some simple facts about this sequences:

1. If nsin(n) converges to the real number a and ncos(n) converges to the real number b, then (n^2)[sin(n)]^2+(n^2)[cos(n)]^2=n^2 converges to a^2+b^2. So the sequences nsin(n), ncos(n) can not converge simultaneously.

Also we have 2nsin(2n)=4nsin(n)cos(n)=(4/n)[nsin(n)][ncos(n)] and

2ncos(2n)=(2/n)[(n^2)(cos(n))^2-(n^2)(sin(n))^2]. Clearly now, 2nsin(2n) and 2ncos(2n) converge to 0, and being subsequences of initial sequences, we have a=0 and b=0.

CONCLUSION: Only one of the sequences nsin(n) and ncos(n) can converge, and in this case, its limit is 0.

2. Using same arguments, we can obtain results about the sequences n·|sin(n)|; n·|cos(n)|.

So it's elementary to prove that all four sequences are not convergent.

Presumably, all four has no limit. But the proofs are not so simple, I think!

Very well documented, Kåre, as usual! Thank you again for your contributions!

Thank you Octav, and keep in touch.

I agree with you, Dinu - we all need more and deeper insight on this issue. I will try to "spread" (to share) this problem throughout Researchgate, to try to engage all its members interested in mathematics.

Octav Olteanu , I like your approach. I thought about a little bit similar reasoning but less strict (in some sense). First, elements of n·sin(n) are switching from positive to negative values and back. Because each "hill" and "valley" on the sinusoid has the width pi, so in each interval corresponding to either one we have 3 or 4 elements of sin(n) (signs of sin(n) : +,+,+, -, -, -, +, +, +, -,-,-,...). We have an infinite subsequence of positive elements and another infinite subsequence of negative elements. The only common limit for both the subsquences could be zero. But, of course it cannot be zero, since in each interval of the length pi at least one sin(n) is "far" from zero (Octav showed a estimation where "far" is sqrt(1/2)). So none of the subsquences can have the limit 0. Multiplying by n doesn't change the sign, so the subsequences of sin(n) which are "far" from zero "transform" into the subsequences of n·sin(n) which tend to plus and minus infinity, respectively. How about the rest of the elements of sin(n) which are closer to zero than sqrt(1/2)? I suppose that after multiplying by n, the remaining n·sin(n) also tend either to plus infinity (positive ones) or minus infinity (negative ones). Unless there is a subsequence of sin(n) which converges to zero fast enough ie. like 1/n^a, a>1. Then, the corresponding n·sin(n) subsequence would converge to zero, too. Is the latter case true? I doubt.

The above consderations also apply to the cos(n) case.

Applying the absolute value on sin or cos removes negative values from the sequences and the minus infinity limit for a subsequence of negative elements does not exist anymore.

Dear Professor Przemyslaw Kowalik,

Thank you for these comments and for your interest in this problem.

Dear contributors, I have searched on internet, and found .... not our sequences, but their inverses. Those thoughts can move us forward:

https://www.quora.com/How-do-I-show-that-the-sequence-1-n-sin-n-diverges

https://math.stackexchange.com/questions/665776/convergence-of-the-sequence-frac1n-sinn

If I am not mistaken, this paper has not been published: https://arxiv.org/pdf/1104.5100.pdf

And yet! https://www.quora.com/Does-the-sequence-n-sin-n-have-a-convergent-subsequence

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

EDIT

I deleted my first answer because

Because of 1. and 2. it became misleading information.

Hello Georges ,

Here some answers on math stackExchange I found

https://math.stackexchange.com/questions/221018/is-n-sin-n-dense-on-the-real-line

https://math.stackexchange.com/questions/1725321/how-would-you-evaluate-liminf-limits-n-to-infty-n-mathopen-sin-n?noredirect=1&lq=1

https://math.stackexchange.com/questions/126583/does-n2-cos-n-diverge-to-infty?noredirect=1&lq=1

Thank you kindly, Moubinool! It looks like the proof of "the limit does not exist" is correct; they prove that liminf n|cos(n)| ≤ π/2 and that limsup n|cos(n)|=+ ∞. I checked the last one and looks correct to me (some of our contributors provided correct arguments as well), and I need to check some details for the first one, involving continued fraction approximations of π (which are not my cup of tea, but "à la guerre comme à la guerre"... All the best, George

Dear George Stoica

I think it is obvious that all the limits

limncos(n), limnsin(n),limn|cos(n)|,limn|sin(n)| does not exist as n goes to infinity.

It is well known that by Bolzano - Weierstrass theorem that:

If the given sequence is bounded, then it should admit a convergent subsequence.

Unfortunately, all the given sequences are unbounded.

In fact, -1 ≤ cos(n) ≤ 1, then -n ≤ ncos(n) ≤ n.

(a similar argument for other sequences)

Concerning, convergent subsequence!

Assume that {nkcos(nk} is convergent for some subsequence {nk} of {n}

where nk goes to infinity as n goes to infinity. (This condition is a must to be a subsequence)

similarly, -nk ≤ nkcos(nk) ≤ nk, which is a contradiction.

(Because every convergent sequence should be bounded.

Here, pi and its approximation didn't play any rule!

Hence, no convergent subsequences exist.

Best regards

Correct, Issam, we need to make the difference between the unbounded ones: not having limit, like (-1)n · n; or having infinite limit, like n or -n.

Dear Issam Kaddoura ,

1. The inequalities -nk ≤ nkcos(nk) ≤ nk do not contradict boundedness.

2. It is shown in the former posts, that the sequences n cos(n) and n sin(n) do not possess even the unbounded limits (neither +oo nor -oo). Therefore most of the followers is now looking for the existence of the limit +oo for the sequences |nsin(n)| and |ncos(n)|. As it was pointed out by Viera Čerňanová , there are possibly some theorems true, of the form "for some sufficiently large a the sequence n^a|sin(n)| tends to infinity". It has been conjectured by me, that it sufficies, that a>1 (referring to some analogy with random sequences "divergent to oo" almost surely). But this is just a conjecture. No answer to the question on the divergence to oo of n^a |sin(n)| or n^a|cos(n)| is known by our community of followers ( yet! ), for any a.

3. \pi DOES matter, since the limit points of sin n [as well as of cos(n)] form the whole interval [-1,-1], in particular 0 is a limit point of the sequences (compare to n^b |cos(\pi n)| which obviously approches oo without doubt for any b>0, since 0 is not a limit point of the sequence |cos(\pi n)| )

Best regards, Joachim

Resume what we clearly have:

Let A={sin(n)}n, B={cos(n)}n. A and B are dense in [-1,1].

For s in (0,1), due to the density, we have sequences of natural numbers (nk)k , (mk)k, (pk)k and (qk)k , so that: sin(nk)-> s, sin(mk)-> -s, cos(pk)-> s, cos(qk)-> -s.

Now we obtain:

1. nksin(nk) has limit +oo

2. mksin(mk) has limit -oo

3. pkcos(pk) has limit +oo

4. qkcos(qk) has limit -oo

5. nk|sin(nk)| has limit +oo

6. pk|cos(pk)| has limit +oo

So clearly n·sin(n) and n·cos(n) have no limit, and n·|sin(n)|, n·|cos(n)| do not converge(have not finite limit).

How do we prove that n·|sin(n)| and n·|cos(n)| have not the limit +oo? That's the question, because I think these two sequences have no limit!

Dear Joachim Domsta

My answer is to tackle the sequences that appear in this question and not the other forms.

limncos(n), limnsin(n),limn|cos(n)|,limn|sin(n)|.

Your note about unboundedness is not clear.

Inequality shows the best we can do to trap the given sequences.

If any of the sequences is bounded, then we can apply Bolzano - Weierstrass theorem to find a convergent subsequence, but this is not the case.

In conclusion, all of the mentioned sequences are not bounded.

Also, pi plays a rule for periodicity at best, it doesn't play any rule to study the convergence. It is enough to use the fact -1 ≤ cos(n) ≤ 1.

Concerning the case: n^b |cos( n)|

If b < 0, then n^b |cos( n)| is a convergent sequence.

This holds for the other forms.

Best regards

Dear Issam,

In this answer I would like to explain only one of our mutual misunderstanding (the other leaving to next posts, please).

Thus, the first step I was a brief response to your words:

" Assume that {nkcos(nk} is convergent for some subsequence {nk} of {n}

where nk goes to infinity as n goes to infinity. (This condition is a must to be a subsequence)

similarly, -nk ≤ nkcos(nk) ≤ nk, which is a contradiction.

(Because every convergent sequence should be bounded."

I have understood them as a proof that the bolded assumption is wrong. Due to this I am again (more directly this time) poiting out, that the inequalities do not contradict the bolded claim.

Of course I admit that you meant something different, but I couldn't guess what particularly.

Best regards, Joachim

Dear contributors, if I am not mistaken, it holds (revise please):

(1)

If 1/(n sin(n)) doesn't converge to 0, then 1/|n sin(n)| doesn't converge to 0.

(2)

from "1/|n sin(n)| doesn't converge to 0" it follows directly that n|sin n| doesn't have a limit.

Therefore it could be enough to prove "1/(n sin(n)) doesn't converge to 0".

Dear Viera Čerňanová

It is about the sequence (1/n)cos(n) and not 1/ncos(n).( case b = -1)

Observe that

-1

That's OK, dear Viera. But: is it easier to prove that 1/(n^a sin(n)) does not converge to zero, for a=1? Note: a proof for a=0.5 was given here by Kåre Olaussen .

Best, Joachim

Dear Joachim Domsta

All real, the bounded and not bounded, sequences are in between

-infinity and + infinity.

The mentioned relation is just to show the best bound we have for the given sequence. ( For absolute values we have 0 and + infinity).

The graph of the sequence oscillates between the first and second bisectors in the plane. And graphically you can observe the unboundedness of the mentioned sequences.

Best regards

The hypothesis proposed by Viera Čerňanová is true for all real numbers a greater or equal than 6.606308. Namely, if μ(π) denotes the irrationality measure of π, then by the first part of Theorem 2 in the paper by Max A. Alekseyev (2011), https://arxiv.org/pdf/1104.5100.pdf, it follows that if μ(π) < 1 + a, then na|sin n| → ∞ as n → ∞. Since the best known upper bound for μ(π) is μ(π) ≤ 7.606308… (V. Kh. Salikhov (2008), On the Irrationality Measure of π,. Russ. Math. Surv. 63(3), 570–572), we have μ(π) < 1 + a for each real number a greater or equal than 6.606309, and hence, for each ≥ 6.606309,

na|sin n| → ∞ as n → ∞.

Analogously, by Alekseyev’s result, it follows that n|sin n|0.15137 → ∞, n1/2|sin n|0.07568 → ∞, and generally, n6.606309b|sin n|b → ∞ for all b > 0, as n → ∞.

Dear Viera Čerňanová

I think you want to study the limit of nsin(n), by studying lim of 1/nsin(n).

But, the second function is not defined over the domain of definition, so we can't discuss the limit of a non-well defined function over the domain.

Best regards

Ok, dear Romeo Mestrovic, but this results do not solve our problem!

Since the irrationality measure of any positive transcendental number is greater or equal than 2, from the second part of Theorem 2 in the paper by Max A. Alekseyev (2011), https://arxiv.org/pdf/1104.5100.pdf, it follows that for all positive real numbers a and b such that a < b, the sequence (na|sin n|b), n = 1, 2, … is not convergent. Namely, there are its subsequence which → ∞, and another its subsequence which is bounded above.

Dear Followers,

I followed and read the article mentioned by Romeo.

I discovered that the proof of the theorem (2) is wrong.

https://arxiv.org/pdf/1104.5100.pdf,

I don't know if this article has a correction.

Best regards

Dear Romeo, please look at my above comment regarding existence of limits for these four sequences. The convegence is completely solved(all four are not convergent).

Dear Issam,

the indicated wrong statement is probably not an error but just a mistake. If I am not wrong - the coorected form which does not destroys the proof should be:

" ...let m be such an integer that |n/π−m| ≤1/2 ..."

Reply if something is missed again (I hope that not).

Best reagards, Joachim

Here is one thing one can say about the sequence for n |sin n| and similar statements should be possible for the other series: The sequence contains infinitely many numbers bounded by some finite constant. Therefore, it does not simply diverge to infinity. The idea for proving the statement is to approximate π by a rational number p/q. Since π is irrational, there is Hurwitz's theorem stating that |π-p/q|

Dear Joachim Domsta

I think your new assumption will not fix the proof of the theorem(2).

Later the author used the assumption m = |n/π| to complete the proof.

Best regards

Dear Issam Kaddoura and Joachim Domsta,

I see that this Alekseyev’s paper is also available at RG, with 3 citations (https://www.researchgate.net/publication/51889294_On_convergence_of_the_Flint_Hills_series/stats). Also, 8 citations of this paper can be found at Alekseyev’s Google Scholar profile (https://scholar.google.com/citations?hl=en&user=1VP4SOMAAAAJ), and two citations (in 2018 and 2019) of this paper among arXiv’s preprints are available at https://ui.adsabs.harvard.edu/abs/2011arXiv1104.5100A/citations (these two arXiv’s preprints are available at https://arxiv.org/abs/1807.02955 and https://arxiv.org/abs/1902.08817). Notice that in this second Carella’s paper (33 pages, selected at arXiv in the “suspicious area/section!!!” GM (General Mathematics), it is proved that the irrationality measure of arbitrary positive irrational number is 2, and hence μ(π) =2. If this result is correct, then this result together with the first part of Theorem 2 in the Alekseyev’s paper implies the truth of hypothesis proposed in the answer by Joachim Domsta (posted 3 days ago; also see my previous two answers), i.e., that the limit relation

na|sin n| → ∞ as n → ∞

is satisfied for arbitrary real number a > 1.

Reminder

Many of the follower's answers are based on the theorem (2) in the article

https://arxiv.org/pdf/1104.5100.pdf,

The author assumed that

(n/π)-⌊(n/π)⌋ ≤ (1/2)

which is the wrong fact.

Counterexample:

(11)/π) - ⌊(11)/π⌋= 0.50141> 1/2.

Best regards

Dear Romeo,

Regardless of the number of citations, we are in process to discuss the correctness of the basic theorem(2) in the article:

https://arxiv.org/pdf/1104.5100.pdf

Best regards

Addendum to my previous post: since successive approximants pi/qi, pi+1/qi+1 of a continued fraction in standard representation satisfy the fundamental identity piqi+1−qipi+1= (−1)i+1, one of the two products must always be odd, so any irrational number has infinitely many approximants with odd denominator.

Dear K. Kassner

( The sequence contains infinitely many numbers bounded by some finite constant. ) How?

f(x) = xsinx has the extremum values at x = π /2 + k π , k is an integer.

And -x ≤ xsinx ≤ x, following the graph of f(x), we deduce that it should diverge. A convergent subsequence is far fetched.

Best regards

A continued fraction is beneficial to approximate functions, numbers, and sometimes to solve diophantine equations in number theory. In this context, it is not clear why we need to use this technique!! I think this technique is useful to study the summation and the convergence of the infinite series, for example, the Flint Hills series, but not the sequences under consideration. Best regards

Issam Kaddoura "( The sequence contains infinitely many numbers bounded by some finite constant. ) How?"

The proof has been sketched in my previous post. Of course, there are also infinitely many numbers in the sequence exceeding any finite bound. You cannot prove anything about this question by just considering the continuous function x sin x. It has infinitely many zeros, so contains infinitely many values bounded by a constant, but the zeros are all irrational numbers and the question is about integer arguments. Equally, the extrema do not occur at integer x (but also not at the places you indicate, they are rather given by a transcendental equation: x=-tan x), so from the fact that their values become arbitrarily large, you cannot directly conclude that the sequence n |sin n| has a divergent subsequence. But I have given the outline of a proof for that as well.

"A convergent subsequence is far fetched."

In fact, a convergent subsequence is not far-fetched but provable. I have shown that there is a bounded subsequence containing infinitely many elements via Hurwitz's theorem. The subsequence has a lower and an upper bound (the upper bound is smaller than 4 and the lower bound is zero, because n |sin n| is non-negative). Then it has an accumulation point in the interval between the two bounds. And there is a theorem (by Bolzano-Weierstraß) saying that there is at least one subsequence (of the subsequence) converging against the accumulation point. (In fact, this is trivial: if you have an accumulation point, you have infinitely many members of the sequence in each small neighborhood of the point. Walk through these members and throw away each of them that is farther from the accumulation point than the previous one considered. Keep the others. There must still be infinitely many, due to the accumulation point property. So you have just constructed a convergent subsequence.)

K. Kassner

I considered The Bolzano- Weierstrass theorem in my previous answers.

The bounded sequence includes a convergent subsequence.

Here, the given sequence is not bounded, then we can't apply this theorem.

In addition, as n goes to infinity, the sequence itself goes to infinity, follows the graph of the function. Where is the assumed accumulation point? I didn't find any connection between the approximation of pi and the given sequences!!

It is another issue. The sequence is considered for n = 1,2,3,4,.....as n goes to infinity. The selected values of nk should be an increasing subset of { 1,2,3,4,...} to construct the subsequence.

To find an approximation of the function nsin(n) in a neighborhood of some point is a different problem.

Since I am the one that asked some of these questions it is better to say something to close the discussion.

Mr. Kassner proof is correct and was known to some people before.

Mr. Kaddoura's arguments are wrong. A subsequence of an unbounded sequence can be bounded. The graph of x sin x is absolutely irrelevant to this question. Why don't you look at the graph of x sin (2 pi x) and at the sequence n sin (2 pi n) and see how they connect.

Also, Mr. Kaddoura's counterexample to Alekseyev's statement does not prove anything. Alekseyev's argument is of the type for all n big enough. If you want to disprove it you need to show that it fails for infinitely many values of n. The fact that it fails for n = 11 is irrelevant.

Few more things about other comments: the fact that cis n is dense in the unit circle is irrelevant. n cis n has limit infinity.

There is usually NO connection between the limit of a sequence and the limit of a function. In Baire category sense there are more functions without limits than functions with limits. For every function f and every points a such that lim f(x) does not exist as x approaches a there are infinitely many sequences x_n with limit a such that f(x_n) is convergent. A more accurate statement will be limits of sequences can say something about limits of functions while limits of functions rarely say something about limits of sequences. Unless one believes that calculus is the real thing.

I will like to point that the proof above while settling the original question did not say much about the behavior of the sequence.

There are bounded subsequences and of course this implies that there are subsequences convergent to a finite limit. However, the real question has become if n |sin n| and n |cos | are dense in [0, infinity)

There are also some similar sequences (subsequences of these) that cannot be proved with a Hurwitz type argument. I will mention some and some related questions. I do know some of the answers but not all.

1. Is sin (n^2) dense in [-1, 1]?

2. Is sin (n^n) dense in [-1. 1]?

3. Is sin (n !) dense in [-1, 1]?

4. Does lim n^2 | sin (n^2) | = infinity?

5. Does lim n^n | sin (n^n) | = infinity?

6. Does lim n ! | sin (n !) | = infinity?

Issam Kaddoura " Here, the given sequence is not bounded, then we can't apply this theorem."

A weird statement. Why don't you simply say you don't understand my proof. Then we can end the discussion.

I have shown, not in much detail, but nonetheless in a way that anyone fluent in mathematics can fill the details in themselves, that the sequence under discussion has a bounded subsequence with infinitely many elements. That was by using Hurwitz's theorem and completed in my first post. The upper bound is at most 4, the lower bound is zero. Now this infinite subsequence must contain, via Bolzano-Weierstraß a convergent subsequence. (A subsubsequence, if you will.) No need to say more.

@ GTP

(*)You are not. George Stoica asked this question.

You said

( A subsequence of an unbounded sequence can be bounded. )

In general, this statement is wrong. So, don't show particular cases.

We discuss the cases nsin(n), ncos(n).

Show me any bounded subsequence if any !!

Your interpretations are strange and wrong!

(*)The mentioned paper includes https://arxiv.org/pdf/1104.5100.pdf, errors to be corrected!! n =11 is to show a counterexample to the proof of Thm(2).

(*) Subsequences and approximations are different issues.

So we have nothing to do with Hurwitz's theorem.

In the end, I am not interested in your new questions.

K. Kassner

My statement is related to the general sequence nsin(n) and not your proposed

bounded subsequence for the pi,'s in ( pi |sin pi|

Gabriel T. Prajitura

If you are not satisfied with n =11;

You can consider

(10000000011/(3.1416))-⌊(10000000011/(3.1416))⌋= 0.88694 >1/2

So, the statement (n/π)-⌊(n/π)⌋ ≤ (1/2)

is wrong for small or large values.

Mr. Kaddoura,

This is a question that was posted here by George Stoica after I refused to do it.

Why did he do it you'll have to ask him.

A subsequence of an unbounded sequence can be bounded. It is a true statement. It cannot be wrong as long as it is a general fact. You confuse unbounded with having infinite limit.

Mr. Kassner showed you a proof of the existence of bounded subsequences. It is not my problem if you can't understand it.

Once again, in Alekseyev's proof he used the statement you consider wrong "for all n big enough" So your counterexample for n = 11 does not show that the proof is wrong, it only shows that that particular statement is wrong for n = 11. In order to show that the proof is wrong you need to show infinitely many values of n for which it is not true. Again, it is not my problem if you don't understand this.

My interpretations are correct statement and it is of no importance to me if you understand them or not. As it is of no importance if you want to look at my questions or not.

Mr. Gabriel T. Prajitura,

(*) If you try a little bit more values for n, you can deduce that it is the wrong relationship (n/π) - ⌊(n/π)⌋ ≤ (1/2) for several large values.

( Once again, in Alekseyev's proof he used the statement you consider wrong "for all n big enough" So your counterexample for n = 11 does not show that the proof is wrong ....)

Recall:

If you are not satisfied with n =11;

You can consider

(10000000011/(3.1416))-⌊(10000000011/(3.1416))⌋= 0.88694 >1/2

So, the statement (n/π) - ⌊(n/π)⌋ ≤ (1/2) is wrong for small or large values.

(*) If you believe ncos(n) admits a convergent subsequence show me one!

It is better to show examples than to judge other's answers!

I agree that the attempt using the approximation Hurwitz's theorem by K. Kassner is an excellent idea. But the details do not guarantee the existence of such subsequences. I assumed a contradiction between the conditions

for the integers pi to satisfy

(1) An increasing subsequence of N.

(2) To satisfy the approximation requirements.

Dear Issam, let me kindly repeat my request to answer to the following explanation. Could you agree, and if not - then why. Thanks in advance :-)

The indicated wrong statement is probably not an error but just a mistake. If I am not wrong - the coorected form which does not destroys the proof should be:

" ...let m be such an integer that |n/π−m| ≤1/2 ..."

Reply if something is missed again (I hope that not).

Best reagards, Joachim

I don't have any particular interest in Alexeyev's proof. My point was that unless you can show a sequence of values of n for which it fails you can't accuse someone of a wrong proof. You may or may not be right, but some numbers, no matter how big, are not enough to contradict that proof. Show a pattern and then you are right.

Getting back to Hurwitz, let us try again.

First, if there is a bounded subsequence there is a convergent subsequence. Mr Bolzano and Mr Weierstrass took care of that.

Second, if there are infinitely many fractions for each possible denominators you can have only a finite number of repetitions. This implies that, once you have infinitely many fractions you also have infinitely many denominators that can be ordered as an increasing subsequence of N. Since the fractions approximate pi an increase in the denominator assures and increase in the numerator.

To Kåre Olaussen:

As noticed in Correction from your answer posted 4 days ago, “I do not make proper use of the Dirichlet theorem above...,” However, computational results sugests that the converse of your inequality |sin(pj , qj)| > 1/qj is satisfied for every j > 4. Namely, the first 11 convergents pj/qj to π are: 3/1, 22/7, 333/106, 355/113, 103993/33102, 104348/33215, 208341/66317, 312689/99532, 833719/265381 (see OEIS Sequences https://oeis.org/A002485 and https://oeis.org/A002486), a calculation in Mathematica 9 gives the following values for the products qj|sin(pj , qj)| (j = 1, 2,...,11):

0.14112, 0.433442, 85.291, 0.384913, 19587.9, 11882.9, 33988.8, 28338.6, 152855.0.

Furtheromore, a calculation in Mathematica 9 (up to n < 108) leads to the following conjecture:

n1/2|sin n| → ∞ as n → ∞.

Dear Joachim,

Thank you for your answer. You agree that there is a mistake, that is (n/π) - ⌊(n/π)⌋ ≤ (1/2) the wrong formula. And we are in the process of fixing that mistake to make the proof details work. You have a suggestion to fix the proof. Can we make a personal correspondence concerning this paper? I appreciate this. Best regards

Mr. Gabriel T. Prajitura,

(*) you can't accuse someone of a wrong proof

I don't accuse anyone, but you accuse my remark about the proof.

You said

(1). ( if there are infinitely many fractions for each possible denominators you can have only a finite number of repetitions.)

(2) This implies that, once you have infinitely many fractions you also have infinitely many denominators that can be ordered as an increasing subsequence of N.

(3) Since the fractions approximate pi an increase in the denominator assures and increase in the numerator.).

We have three claims or conjectures. Each one needs proof. So, we return to the first square and nothing can be guaranteed. It is better to seek an example if exist!

Romeo Meštrović "As noticed in Correction from your answer posted 4 days ago, “I do not make proper use of the Dirichlet theorem above...,”

Yes, by trying to do calculations in my head I made a really silly mistake. To repeat the correct argument: The Dirichlet approximation theorem says that for every irrational number x there exist an infinite sequence of rational numbers {pj/qj} (with pj, qj integers) such that |x - pj/qj | < 1/qj2. Since π is an irrational number this means that there is an infinite sequence { pj } of integers such that

pj = qj π + εj, with qj integer and |εj| < 1/qj (this is the relation I first wrote wrong).

It follows that pj |sin pj | = pj |sin εj | < pj |εj | < pj/qj < 4 for infinitely many pj. So the sequence {n | sin n |} has an infinite bounded subsequence. Since it also has infinite unbounded subsequences, it does not have any finite or infinite limit.

So, you should look ay the sequence {pj | sin pj |}, not the sequence I mistakenly gave first. The mistake was actually corrected in the added note at the bottom of my post(s).

Dear Issam Kaddoura,

The sequence a(n)=(1+(-1)^n)sin(n)+(1+(-1)^(n+1))n is unbounded. The subsequence a(2n) of a(n) is bounded, and consequently a(2n) contains a convergent subsequence.

If you wish "see" effectively a convergent subsequence of an unbounded sequence, we can consider b(n)=(1+(-1)^n)sin((n π )/4)+(1+(-1)^(n+1))n, which is unbounded, b(2n) is a bounded subsequence and b(4n)=0 for all n, being so a convergent subsequence of b(n) and b(2n).

The proof of K. Kasner is very clear, n |sin(n)| contains a convergent subsequence and from the fact that {sin(n)} is dense in [-1,1], it results in a simple approach, that n |sin(n)| has a subsequence diverging to +infinity. Consequently n |sin(n)| has no limit.

Dear Issam and Joachim,

The error in Alekseyev’s proof, observed by Issam Kadoura, is irreparable and related false statement is subsantial in the proof of the first part of Theorem 2 (page 2 of the paper). Namely, as noticed by Issam, the inequality |n/ π –[n/ π]| < 1/2 is not generally true. In fact, a computation suggests that this inequality does not hold for infinitely many values of n; namely, for numerous values of n, |n/π –[n/π]| is close to 1. However, |n/π –[n/π]| < 1/2 is essential in the rest of the proof, in view of the fact that the upper bound ½ gives the possibility to apply Lemma 1 (i.e., the well known inequality |sin x| ≥ 2|x|/π when |x| ≤ π/2).

Best regards

Romeo Meštrović "Furthermore, a calculation in Mathematica 9 (up to n < 108) leads to the following conjecture: n1/2|sin n| → ∞ as n → ∞"

Did you look carefully at every term in that sequence? Here is the first values of p | sin p |, where p is the numerator of the continued fraction approximants; they do indeed satisfy the inequality of my previous posts

22: 0.195

333: 2.937

355: 0.011

103993: 1.989

104348: 1.149

208341: 1.691

312689: 0.907

833719: 1.928

Mr. Kaddoura,

I did not accuse you of anything. I said you may be right or wrong, it just does not follow from your argument.

Mr. Teodorescu,

Good luck with that. At this point I give up.

Dear Romeo Meštrović

Thank you for the calculations. I told Joachim Domsta in my answer 6 hrs ago that the author used this fact powerfully in the rest of the proof.

I am in process to send him more details. Best regards

Dear Dinu Teodorescu

(*) I think you get me wrong.

I didn't say that unbounded sequences don't include bounded subsequences!!

Ex : x(n) = ( 1 + (-1)n )n is divergent unbounded equence, but

x(2n+1) = x(4n+1)=x(6n+1)=....= 0, convergent bounded subsequences.

My statement was for the sequences under consideration.

(**)

Your statement

(that n |sin(n)| has a subsequence diverging to +infinity. Consequently n |sin(n)| has no limit.)

This statement supports my argument and not K. Kasner proof!

Best wishes

Dear Issam and Joachim,

Namely, if in Alekseyev’s proof we replace his wrong inequality |n/ π –[n/ π]| < 1/2 by the inequality |n/ π –[n/π]| < 1 (which is true by definition of Floor function [ ]), then in order to apply Lemma 1, if we put m = [n/π], it is necessary the previous inequality to write in the form |n/ 2– m/2| < π/2, i.e. (since the inequality |sin x| ≥ 2|x|/π of Lemma 1 is satisfied when |x| ≤ π/2, but obviously not in the range |x| ≤ π!). However, the first equality |sin n| = |sin (n - mπ)| in the chain of equalities/inequality on the last line on page 2, cannot be replaced by the equality |sin n/2| = |sin (n/2 – mπ/2)| becuse the previous equality holds only for even values of m. Finally, the rest of the proof of the second part of Theorem 2 is based on the inequality obtained on the basis of identity |sin n| = |sin (n - mπ)|, which now is not generally applicable with m/2 instead of m!.

Accordingly to the previous mentioned, and taking into account the whole proof of the firdt part of Theorem2, it seems that the first assertion of Theorem 2 is true for the subsequence nk of natural numbers such that [nk/π] is even number.

Best regards,

Romeo

Romeo Meštrović

Your remarks are constructive to improve the mentioned article.

I think we have shared ideas that may fix the Thm proof.

Best wishes

To Kåre Olaussen:

Thank you for clarifying your proof. Clearly, using the idea of your proof and the concept of irrationality measure of π (μ(π)), if μ(π) > 2 then your result can be extended as follows.

For any arbitrarily small real number ε > 0, put a = μ(π) – ε -1. Then there exist infinitely many positive integers pi (i = 1, 2,…) such that the sequence {pii^a|sin pi|} is bounded (in fact, by a constant 4^a).

Also my “small” calculations led to the erroneous hypothesis about the limit of the sequence {n1/2|sin n|}.

https://math.stackexchange.com/questions/309680/limit-computation-lim-n-to-infty-left-sin-n-right-n-infty?noredirect=1&lq=1

https://math.stackexchange.com/questions/221018/is-n-sin-n-dense-on-the-real-line

At this point, I am sending my thanks to all contributors. Special thanks to Klaus Kassner, Kåre Olaussen and Romeo Meštrović, whose arguments I followed and found correct: our sequences (with absolute value) have no limit.

However, I will leave this track of thought open, to allow our contributors to (eventually) settle among themselves: extensions, related topics, etc.

The idea from very elegant complex proof posted by George Stoica (July 4) that the (finite) limit of the sequence {nsin n} does not exist as n → ∞, can be used for the short elementary proof of this fact (based on the addition formula for the sine) given as follows. Assuming that nsin n → L, with L < ∞, we have (n + 1)sin(n +1) → L, i.e., nsin(n +1) + sin(n +1) → L. As (n +1)/n → 1, it follows that sin(n +1) → 0, i.e., sin n cos 1 + sin 1cos n → 0, whence we get cos n → 0. Hence, we have 1 = cos^2 n + sin^2 n → 0 as n → ∞, i.e., 1 = 0. A contradiction.

Similarly, from (n + 1)cos(n +1) = ncos(n +1) + cos(n +1) and the identity cos(n +1) = cos n cos 1 - sin n sin 1, we can deduce that the (finite) limit of the sequence {ncos n} does not exist as n → ∞.

Mr. Mestrovic,

if n sin has limit L then n | sin n | has limit | L | so for n big enough

| sin n | < ( | L | + 1) / n which implies sin n has limit 0

To Gabriel:

I agree, but the sequence {nsin n} is not convergent. In particular, your assertion is true for all its convergent subsequences (which exist because there are its bounded subsequences), as well as for its subsequences that → ∞.

The statement is far more general

If x_n y_n has finite limit and x_n has infinite limit then y_n has limit 0 while if x_n is unbounded y_n has a subsequence with limit 0.

In the case of n sin n it just provides a simple proof that the limit is not finite which is not that hard anyway.

What I find fascinating about this sequence, sin n, is that it has subsequences converging to any number in [-1, 1] but if you multiply it by n it gets subsequences convergent to any real number and +/- infinity. All this subsequences to infinite limits come from subsequences converging to numbers in [-1, 0) union (0, 1] while the little subsequence (it has density 0) with limit 0 is responsible for all finite limits of n sin n

In a similar manner one can prove something similar for [n pi] sin ([n pi] ) or [n pi] {n pi}

The sequences given by George have subsequences that converge to different limits including infinite limits and therefore are not convergent sequences. A sequence converges only when it has unique limit and all subsequences converge to the same limit.

It is interesting to look at this question in a slightly different way:

For which subsequences n=n_j does n sin n converge?

Clearly this is only possible if sin n_j converges to 0 (at least as rapidly as 1/n_j)

which means n_j is close to a multiple of \pi, so we are asking for rational sequences n/m with |n/m-\pi|

Dear Thomas,

I doubt that it is possible to find all such subsequences. If you have two subsequences convergent to the same limit any infinite subset of their union will produce another subsequence converging to that limit.

You are probably thinking at some specific subsequences that we can control well enough so we can say what the limit is.

One step toward this (because we want sin n to get close to 0) is to consider

[n pi] sin ([n pi] ) or ([n pi] + 1) sin ([n pi] + 1) rather than n sin n because these are the closest integers to n pi so the ones for which sin is smaller.

A graph of these two subsequences suggested that their behavior is as bad as for n sin n.

Thus maybe some subsequences of these two will be more illuminating.

However, I do not know if the theory of rational approximation is up to this task yet.

i accepted but the sequesnce is not convergence

nsinn etc

My idea was to change the context slightly and ask about the possible subsequential limits. In that context I will make two conjectures:

[C1] There is some subsequence n=n_j\to\infty such that n sin n converges to 0

I fairly strongly believe this is true, but do not consider it obvious. A somewhat stronger conjecture would be:

[C2] For each real L there is some subsequence such that n sin n converges to L

The question (difficult) would then be whether it makes a difference whether the possible limit would be 0 or not --- possibly whether L>0 or not. The same questions could also be raised for n cos n, but I think it is easy to see that the answers would be the same.

Thomas, I believe that too and other people expressed the same opinion here.

If you look at other answers you can see a link to stack exchange posted by Mr. Omarjee where there was at some point a discussion on this subject.

If n is fractional, you get the sinc function. Does that help?

It is correct Mark, but it opens another problem, beyond sequences and their subsequences.