A Difficult sum

Nice.. but why have you become the "Devil"?

@ 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]

At 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.

any book reference?

I 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

Attempt:

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.

What is the final solution ?

You 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

Well 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.

I 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 {a_n}, {b_n} such for some N we have a_n>b_n

for all n>N, at most we can conclude that lim_{n→∞} a_n ≥ lim_{n→∞} b_n. Strict inequality does not necessarily follow

e.g. {2/n}, {1/n}

Could you please share the fun way of evaluating that.

Instead 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.

No...that isn't mentioned, infact they haven't used it in the proof either...[12]

Oh yes. Π>θ>0