# Prove that the series diverges

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.

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.
• 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. – marty cohen Sep 25 '19 at 20:56