# Determine whether the series$\sum_{n=1}^\infty \frac{\sin n \sin n^2}{n}$is convergent or divergent.

Determine whether the series $$\sum_{n=1}^\infty \frac{\sin n \sin n^2}{n}$$ is convergent or divergent. To apply Dirichlet test, I need to prove $\sum_{n=1}^N \sin n \sin n^2$ is bounded. Is that true? If so, how can I show that?

Both the series $$\sum_{n\geq 1}\frac{\cos(n^2\pm n)}{n}=\text{Re}\sum_{n\geq 1}\frac{\exp\left[(n^2\pm n)i\right]}{n}$$ are conditionally convergent by Weyl's inequality and summation by parts. See also this similar question. On the other hand the partial sums of $\cos(n^2\pm n)$ are not bounded: one may easily adapt the following lines.
Van Der Corput's trick for lower bounds. The purpose of this paragraph is to prove that the partial sums of the sequence $\{\sin(n^2)\}_{n\geq 1}$ are not bounded. We have: \begin{eqnarray*}2\left(\sum_{k=1}^{n}\sin(k^2)\right)^2 &=& \sum_{j,k=1}^{n}\cos(j^2-k^2)-\sum_{j,k=1}^{n}\cos(j^2+k^2)\\ &=&n+2\sum_{m=1}^{n^2-1}d_1(m)\cos(m)-2\sum_{m=2}^{2n^2}d_2(m)\cos m\end{eqnarray*} where $d_1(m)$ accounts for the number of ways to write $m$ as $j^2-k^2$ with $1\leq k<j\leq n$ and $d_2(m)$ accounts for the number of ways to write $m$ as $j^2+k^2$ with $1\leq j,k\leq n$. Since both these arithmetic functions do not deviate much from their average order (by Dirichlet's hyperbola method $d_1(m)$ behaves on average like $\log m$ and $d_2(m)$ behaves on average like $\frac{\pi}{4}$), it is not terribly difficult to prove that for infinitely many $n$s $$\left|\sum_{k=1}^{n}\sin(k^2)\right|\geq C\sqrt{n}$$ holds for some absolute constant $C\approx \frac{1}{\sqrt{2}}$ through summation by parts and the Cauchy-Schwarz inequality. A detailed exposition on Dirichlet's hyperbola method can be found on Terence Tao's blog. We recall that Van Der Corput's trick is usually employed to produce upper bounds: for instance $$\left|\sum_{k=1}^{n}\sin(k^2)\right|\leq D\sqrt{n}\log n,$$ for some absolute constant $D>0$, holds for any $n$ large enough. In particular $\sum_{n\geq 1}\frac{\sin(n^2)}{n^\alpha}$ is convergent for any $\alpha>\tfrac{1}{2}$.