How to prove that $\sum_{n=1}^\infty\frac{\sin n\cdot\sin n^2}{n}$ converges?

Question

The series is $$\sum_{n=1}^\infty\frac{\sin n\cdot\sin n^2}{n}$$ It seems to use Dirichlet's test, but I cannot prove $\sum\sin n\cdot\sin n^2$ is bounded. This question may help -- Convergence of $\sum \limits_{n=1}^{\infty}\sin(n^k)/n$

Answer 1

Approach 1: Telescoping Sum $$ \begin{align} \sum_{k=1}^n\sin(k)\sin\left(k^2\right) &=\frac12\sum_{k=1}^n(\cos(k(k-1))-\cos(k(k+1)))\\ &=\frac{1-\cos(n(n+1))}2\tag1 \end{align} $$ Thus, the partial sums are bounded by $1$. Dirichlet's Test and $(1)$ say that the original series converges.

---

Approach 2: Summation by Parts $$ \begin{align} \sum_{k=1}^\infty\frac{\sin(k)\sin\left(k^2\right)}k &=\lim_{n\to\infty}\sum_{k=1}^n\frac{\sin(k)\sin\left(k^2\right)}k\\ &=\lim_{n\to\infty}\frac12\sum_{k=1}^n\frac{\cos(k(k-1))-\cos(k(k+1))}k\\ &=\lim_{n\to\infty}\left(\frac12-\frac{\cos(n(n+1))}{2n}-\frac12\sum_{k=1}^{n-1}\frac{\cos(k(k+1))}{k(k+1)}\right)\\ &=\frac12-\frac12\sum_{k=1}^\infty\frac{\cos(k(k+1))}{k(k+1)}\tag2 \end{align} $$ and the last sum converges by comparison to $$ \sum_{k=1}^\infty\frac1{k(k+1)}=1\tag3 $$