Diverge? $\lim_{n \to \infty} \sin(n)+ \sin(n^2)=?$ I encounter with a problem that is to find:
$$\lim_{n \to \infty, n \in \mathbb{N}} \sin(n)+ \sin(n^2)=?$$
So I do know the set $\{\sin n: n \in \mathbb{N}\}$ is dense in $[-1,1]$ and also knowing that $\lim_{n \to \infty} \sin(n^2) $ does not converge and it even has infinite limit points. However, I'm running in a trouble that is proving that if $\sin(n)+ \sin(n^2)$ converges, and exists such a subsequence $\{n_k\}_{k\geq 1} \subset \mathbb{N}$,  $\lim_{k \to \infty} \sin(n_k^2) = a$ then: $$\lim_{k \to \infty} \sin(n_k) \neq L - a$$.
 A: If we assume the results mentioned in Terence Tao's answer to a related question, there's an easy proof that the limit does not exist.
Pick some suitably large natural number $ N $ and define the random variables $ X(k) = \sin(k) $, $ Y(k) = \sin(k^2) $ on the set of positive integers from $ 0 $ to $ N-1 $ seen as a probability space with the uniform measure. Tao's results say that for "almost all" values of $ N $ (in the sense of natural density, for instance) we have the bounds $ \mathbb E[X] = \mathbb E[Y] = o(1) $ as well as $ \mathbb E[X^2] = \mathbb E[Y^2] = 1/2 + o(1) $. (Here I don't mean that the expectations over $ X $ and $ Y $ are equal, just that they are both in the same asymptotic class.) Suppose we pick such an $ N $. In this case, we have that
$$ \mathbb E[(X+Y)^2] = \mathbb E[X^2] + \mathbb E[Y^2] + 2 \mathbb E[XY] = 1 + \frac{2}{N} \sum_{k=0}^{N-1} \sin(k) \sin(k^2) + o(1) $$
$$ = 1 + \frac{1}{N} \sum_{k=0}^{N-1} (\cos(k^2 - k) - \cos(k^2 + k)) + o(1) $$
The partial sums of both $ \cos(k^2 - k) $ and $ \cos(k^2 + k) $ are also $ O(\sqrt{N}) $ by the same result of Tao, and so we find that $ \mathbb E[(X+Y)^2] = 1 + o(1) $ for most values of $ N $. On the other hand, we also know that $ \mathbb E[X+Y] = o(1) $ for most values of $ N $. No convergent sequence can have both of these properties at once, so we conclude that $ \sin(n) + \sin(n^2) $ does not converge.
