Prove this :
\lim _{n\rightarrow \infty } \frac{|\sin \theta|+|\sin 2\theta|+...|\sin n\theta|}{n}>0
I must admit that I've its soln. and I'd not have been able to solve it by myself otherwise. But I liked the sum, so flicked it here :D
341I get that it is a strict inequality between the two sequences.
The same holds for the sequences I provided in #12.
Yet you cannot conclude that \lim_{n \rightarrow \infty} \frac{2}{n}>0 when in fact its zero.
The point is that a sequence may have all members positive and yet end up with limit zero. You use the same argument to state that a_n>b_n may hold for all n, yet their limits can be equal
As for the discussion on the integral, we were done with discussing its divergence quite some time ago when Ricky posted this integral as a question. So we can stop that discussion (in any case i did mention that I was doing some hand waving.)
341I was reminded of a beautiful inequality I had read about. It's called Fejer's Inequality.
It states that if 0< \theta < \pi, then
\sum_{k=1}^n \frac{\sin k \theta}{k}>0
for any natural number n.
Here's one reference I could find http://www.emis.de/journals/AMI/2006/koumandos.pdf
11@ Kaymant sir, I too thought of that at first sight, but then got carried away seeing the soln. at hand [3]
This is the outline of the solution:
Lemma Used - If |sinmθ|≤sinε for some m, then |sin(m+1)θ|>sinε
Using this, we conclude that at least n-12 of the summands in the numerator are greater than sinε
Thus \lim _ {n\rightarrow \infty} \frac{\sum_{k=1}^{k=n} |\sin k\theta|}{n}\ge \lim _{n\rightarrow \infty} \frac{1}{n}.\frac{n-1}{2}\sin \epsilon =\frac{\sin \epsilon}{2}>0
[1]
66At some point of time I was thinking in that direction, specifically of the Parseval's theorem but did not give much energy to it. Probably now some alternate proof (apart from that given by Bhatt sir) could be thought of.
11@ bhat sir,
This was actually from a handout I downloaded some days back from the net. It was actually a paper over Fourier Series (that I needed for my project), and they used this result in some proof. This was actually proved as a theorem. [1]
11H'mm...so that is another proof to this qsn...[1]
I'll post the soln. given, later as it's pretty long, and now I've two assignments to complete [3]
341soumik, I am fairly certain of my proof. can you pitch in and put us out of our misery by (in) validating it and/or providing the solution?
341I am kind of unhappy with that solution. So let me modify it a bit
The basis is the same, that the average of n numbers is bounded below by the least of the numbers.
Now let \theta = t \pi, 0<t<1. Suppose further that \frac{1}{10^a}<t \le \frac{1}{10^{a-1}}
c = \frac{1}{10^a}
Its easy to see that among the numbers kt, k =1,2,...,n, we will have
\{kt\} < c at most \frac{n}{2} times (when t=0.5). Let k not satisfying this inequality belong to a set denoted by \mathbb{S}. Let \left|\mathbb{S}\right| = m
That means if we k \in \mathbb{S}, then we have \sin k \theta \ge c>0
Now, we can write
LHS \ge \frac{\sum_{k=1, k \in \mathbb{S}}^n \sin k \theta}{n} = \frac{\sum_{k=1, k \in \mathbb{S}}^n \sin k \theta}{m} \times \frac{m}{n}
Now, the first factor ≥ sin c
And the second factor ≥ 1/2
Hence the limit is at least \frac{\sin c}{2}
which is positive
341Attempt:
A simple argument would be based on the fact that the average of n numbers is greater than or equal to the least of them
Case 1: θ is an irrational multiple of π, then of course none of the numbers,\sin k \theta are zero.
Then the average of these numbers will be greater than or equal to the least of the numbers which is itself greater than zero.
Case 2: \theta = \frac {p \pi}{q} with p, q \in \mathbb{Z^{+}} and gcd (p,q)=1
Then all \sin k \theta where q|k equal zero. And the rest are non-zero.
Now we may assume that n is a multiple of q (and use sandwich theorem for completing the argument. Or say that the limit remains unchanged when we ignore finitely many terms as the number of omitted terms<q).
So let n = mq.
Then the given sum can be written as
\frac{\sum_{k=1, q \not | k}^{n} |sin k \theta|}{n-m} \ \times \ \frac{n-m}{n}
As in the previous argument, the first factor>0
And the second factor is \left(1-\frac{1}{q} \right)>0
Hence the limit>0
Note that this argument totally fails when the lower bound itself can be brought arbitrarily close to zero.
341You need some condition on θ too. If \theta = n \pi,for example, the limit is zero. Maybe you want \theta to be an irrational multiple of \pi
66Well actually I was using strict inequality because the condition for equality in triangles inequality is not met here. In case of triangles inequality, |x1 + x2| ≤ |x1| + |x2| equality holds when x1 and x2 are of the same sign. In this case, for sufficiently large n not all sines will have the same sign. And that's why I took the strict inequality.
As for the integral, the function I(a) taken is not continuous at a=0, and so I(0) ≠1. In fact,
I(a)=\int_0^\infty e^{-ax}\sin x\ \mathrm dx
converges only for a>0.
341As for the integral
http://www.thestudentroom.co.uk/showthread.php?t=704841
341I did try the triangle inequality and evaluating the sum as you have done. But the conclusion does not follow from this.
If we have two sequences {an}, {bn} such for some N we have an>bn
for all n>N, at most we can conclude that limn→∞ an ≥ limn→∞ bn. Strict inequality does not necessarily follow
e.g. {2/n}, {1/n}
66Instead the triangles inequality suffices.
\sum_{k=1}^{n}|\sin k\theta| > \left|\sum_{k=1}^n\sin k\theta\right|
The sum within the modulus comes out to be
\dfrac{\sin \theta -\sin(n+1)\theta +\sin n\theta}{2(1-\cos\theta)}
Now dividing by n, using the boundedness of sin and taking the limit, we get 0 and the result follows.
341Hand waving: We can show that this is greater than 0∫∞ sin x dx =1.
That integral may be suspect though
341coz its not needed as i realized later. the problem statement looks fine now
11No...that isn't mentioned, infact they haven't used it in the proof either...[12]
341But you still need \theta to be an irrational multiple of \pi. Is that condition mentioned
