The series is: $$\sum_{n=1}^\infty \frac{\cos(\sqrt{n})}{n}$$ Considering it isn't always positive, I replace $\frac{\cos{\sqrt{n}}}{n}$ with its absolute value and I find that: $$\vert \frac{\cos{\sqrt{n}}}{n}\vert\gt \frac{\cos^2{\sqrt{n}}}{n}=\frac{\ 1+\cos{2\sqrt{n}}}{2n}=\frac{1}{2n}+\frac...