Conditional Convergence of a Series $\sum_{n=1}^{\infty}\frac{\sin(n)\sin(n^2)}{n}$ I wanted to show that the series $\sum_{n=1}^{\infty}\frac{\sin(n)\sin(n^2)}{n}$ is absolutely convergent or not. My claim is it is conditionally convergent, to show that I used Dirichlet test and show that the original series is convergent. My question is the following;
Let $g(n)$ be the function from $\mathbb{N}$ to $\mathbb{R}$ as $g(n) = \left|\sin(n)\sin(n^2)\right|$. I want to show that there are at least one finite sequence $\{n,n+1,n+2,...,n+k\}$ satisfying $$g(n) > C \ \lor g(n+1) > C \ \lor \cdots \lor g(n+k) > C \ \forall n\in\mathbb{N} \text{ for some } C>0.$$ To proceed, let us define a set $$K = \{k:|\sin(k)\sin(k^2)| > C\}$$ then we have $$\sum_{n=1}^{\infty}\frac{|\sin(n)\sin(n^2)|}{n} \geq \sum_{n\in K}\frac{|\sin(n)\sin(n^2)|}{n} \geq \sum_{a=1}^\infty\frac{C}{ak}$$ and since RHS is divergent by general harmonic series test, we're done.
But I cannot prove there exists such a sequence, I afraid the solution I wanted to create so difficult also. However, I work on this problem like 3 weeks and I couldn't figure out better way to attack this problem. Any help would appreciated, thanks in advance.
 A: I have solved the question via greater tools than a sophomore can have. I don't believe I can prove the two tool I have used. So I used references to get them.
Convergence of original series is from Dirichlet test. Let $a_n= \frac{1}{n}$, $b_n = \sin(n)\sin(n^2)$ and $c_n = \cos(n^2-n)$. Since $\cos$ is even function we have $\cos(x-x^2) = \cos(x^2-x)$. 
Observe that $$b_n = \frac{\cos(n^2-n)-\cos(n^2+n)}{2} = \frac{c_n-c_{n+1}}{2}$$ So for any $N\in\mathbb{N}$ we have
$$\left|\sum_{n=1}^{N}b_n\right| = \left|\frac{c_1-c_{N+1}}{2}\right| = \left|\frac{1-\cos(N^2+N)}{2}\right| <2.$$
Since we have the necessary conditions to apply Dirichlet test, series converges.
To proceed to show the series is not absolutely convergent, we need to assume the following:
$n$ is equidistributed $\mod \pi$.
$n^2$ is equidistributed $\mod \pi$.
The proofs are immensely hard for me but Corollary 6 of Terrence Tao's work on Equidistribution of polynomial sequences in tori [1] implies both of them. [2] 
We know that $$*|\sin(n^2)| > \frac{1}{2} \text{ whenever } n^2mod\pi\in(\pi/6,5\pi/6) \text{ and} $$ $$**|\sin(n)| > \frac{1}{2} \text{ whenever } nmod\pi\in(\pi/6,5\pi/6).$$ Since both of them are equidistributed for sufficiently large $N\in\mathbb{N}$, for any sequence $\{N+1,N+2,...,2N\}$ we have asymptotically $2/3$ of them are appropriate for * and $2/3$ of them are appropriate for **. Hence by Pigeonhole principle, we have that at least $1/3$ of them satisfying both. Meaning for any sequence $\{N+1,N+2,...,2N\}$ while $N$ is sufficiently large, we have $$|\sin(n)||\sin(n^2)|>\frac{1}{4}$$ holds for at least the $1/3$ of the sequence. Therefore for large enough $N\in\mathbb{N}$ $$\sum_{n=N+1}^{2N}\frac{|\sin(n)||\sin(n^2)|}{n} > \frac{1}{2N}\sum_{n=N+1}^{2N}|\sin(n)||\sin(n^2)| > \frac{1}{2N}\cdot\frac{N}{3}\cdot\frac{1}{4} = \frac{1}{24}$$ Therefore the series diverges since the tail of the series diverges.
$$\sum_{n=N+1}^{\infty}\frac{|\sin(n)||\sin(n^2)|}{n} > \sum_{k=0}^{\infty}\left(\sum_{n=2^k(N+1)}^{2^{k+1}N}\frac{|\sin(n)||\sin(n^2)|}{n}\right)>\sum_{k=0}^{\infty}\frac{1}{24}$$ Our series is not absolutely convergent therefore is conditionally convergent.
Terrence Tao, 2010, accessed 24 April 2020 ,
https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/
Alexis Olson, 2016, accessed 24 April 2020 , Convergence of $\sum_n \frac{|\sin(n^2)|}{n}$
