# Does $\sum_{n=1}^\infty\sin(n)/\sqrt n$ converge absolutely?

So I already proved that $$\sum_{n=1}^\infty \frac{\sin(n)}{\sqrt n}$$ conditionally converges using the Dirichlet test.

I'm almost sure it doesn't converges absolutely, but I'm struggling with proving it using cauchy condition or with other convergence tests.

Since $$\frac{\sin^2n}{\sqrt{n}} \leqslant \frac{|\sin n|}{\sqrt{n}}$$ for any $$n \in \mathbb{N}^{\times}$$, assuming that $$\displaystyle \sum_{n＝1}^{\infty} \frac{|\sin n|}{\sqrt{n}} < \infty$$ would entail that $$\displaystyle \sum_{n＝1}^{\infty} \frac{\sin^2n}{\sqrt{n}} < \infty$$.
However, as $$\sin^2n＝\frac{1-\cos(2n)}{2}$$ one notices that: $$\displaystyle \sum_{n＝1}^{\infty} \frac{\sin^2n}{\sqrt{n}}＝\displaystyle\sum_{n＝1}^{\infty}\frac{1}{2\sqrt{n}}－\displaystyle\sum_{n＝1}^{\infty}\frac{\cos(2n)}{2\sqrt{n}}.$$ Can you see why this would end up entailing a contradiction?