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I want to check whether $$ \sum_{n=1}^{\infty} \frac{\sin n\theta \sin \sqrt{n}}{n} $$ is convergent or not. $\theta$ is a real number. What I know is $$ |\sum_{n=1}^{N}\sin n\theta| = |\frac{\cos \frac{\theta}{2} - \cos(N+\frac{1}{2})\theta}{2\sin\frac{\theta}{2}}| \leq \frac{1}{|\sin \frac{\theta}{2}|}$$ for $\theta\neq 2k\pi$. Let's assume $\theta \neq 2k\pi$. So by Dirichlet test, $\sum_{n=1}^{\infty} \frac{\sin n\theta}{n}$ is convergent. But I don't quite know how to solve the original one. Any hint or something? Thank you so much!

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We may use a combination of summation by parts and the Denjoy-Koksma inequality, since $\sin\sqrt{n}$ is approximately constant on short intervals.
Clearly if we manage to prove that $\sum_{n\geq 1}\frac{1}{n}\cos(n\theta\pm\sqrt{n})$ is convergent/divergent we are done. Let $\left\{\frac{p_m}{q_m}\right\}_{m\geq 1}$ be the sequence of convergents of the continued fraction of $\pi$. $\cos x$ is Lipschitz-continuous and $\left|\pi q_m-p_m\right|\leq\frac{1}{q_m}$, hence $$ \sum_{n=1}^{p_m}\frac{\cos(n\theta\pm\sqrt{n})}{n}=H_{p_m}-\sum_{n=1}^{p_m}\frac{1-\cos(n\theta+\sqrt{n})}{n} $$ can be effectively approximated by $$ H_{p_m}-\int_{0}^{\pi q_m}\frac{1-\cos(\theta x+\sqrt{x})}{x}\,dx = H_{p_m}-4\int_{0}^{\sqrt{\pi q_m}}\frac{\sin^2\left(\frac{\theta}{4} x^2+\frac{1}{2}x\right)}{x}\,dx$$ due to the Denjoy-Koksma inequality. Invoking the Laplace transform, the problem boils down to checking the convergence of the usual Fresnel integrals and their squares. In a more elementary way, $\sin^2\left(\frac{\theta}{4}x^2+\frac{1}{2}x\right)$ has mean value convergent to $\frac{1}{2}$ on long intervals, hence the singular part of the last integral is $$ 4\cdot \frac{1}{2}\log\sqrt{\pi q_m} = \log(\pi q_m)$$ cancelling the singular part of $H_{p_m}$, i.e. $\log(p_m)$. This should be enough to state the convergence of the original series, at least for $\theta\not\in\pi\mathbb{Z}$. If $\theta=0$ we have an interesting sub-question. Since $$ \sum_{n=p_m}^{p_{m+1}}\frac{\sin\sqrt{n}}{n}\approx \int_{p_m}^{p_{m+1}}\frac{\sin\sqrt{x}}{x}\,dx $$ and $\int_{0}^{+\infty}\frac{\sin\sqrt{x}}{x}\,dx=\pi$, it is natural to wonder about a closed form for $$ \pi-\sum_{n\geq 1}\frac{\sin\sqrt{n}}{n}.$$

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