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I am struggling trying to prove that the following series diverges: $$ \sum_{n=1}^{\infty}\frac{\sin{n}}{\sqrt{n}+\sin{n}} $$ I would be very grateful if anyone could give me some clue.

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Well, $\sum_{n\geq 1}\frac{\sin n}{\sqrt{n}}$ is convergent by Dirichlet's test: $\sin(n)$ has bounded partial sums and $\frac{1}{\sqrt{n}}$ is decreasing to zero. We have $$ \frac{\sin n}{\sqrt{n}}-\frac{\sin n}{\sqrt{n}+\sin n} = \frac{\sin^2(n)}{n+\sqrt{n}\sin(n)} $$ and it is enough to show that $$ \sum_{n\geq 2019}\frac{\sin^2(n)}{n+\sqrt{n}\sin(n)} $$ is divergent. Due to the equidistribution of $e^{in}$ in $S^1$ we have that for at least $N-O(1)$ integers in the interval $[N,3N]$ the inequality $\sin^2(n)\geq \frac{1}{2}$ holds, so for any $N$ large enough $$ \sum_{n=N}^{3N}\frac{\sin^2(n)}{n+\sqrt{n}\sin(n)} \geq \frac{1}{2}\cdot\frac{N-O(1)}{3N+\sqrt{3N}}\geq \frac{1}{7} $$ holds and we are done.

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  • $\begingroup$ You can also show that $\sum_{n\geq 2019}\frac{\sin^2(n)}{n+\sqrt{n}\sin(n)} $ diverges by noting that $\frac{\sin^2(n)}{n}-\frac{\sin^2(n)}{n+\sqrt{n}\sin(n)} =\frac{\sqrt{n}\sin^3(n)}{n(n+\sqrt{n}\sin(n))} $ so that $\sum_{n\geq 2019}\frac{\sin^2(n)}{n+\sqrt{n}\sin(n)} =\sum_{n\geq 2019}\frac{\sin^2(n)}{n} -\sum_{n\geq 2019}\frac{\sqrt{n}\sin^3(n)}{n(n+\sqrt{n}\sin(n))} $ and the first sum diverges and the second converges. $\endgroup$ – marty cohen Sep 25 '19 at 20:56

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