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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.

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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?

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