Determine the convergence of $ \sum_{n=1}^{\infty}\left[1-\cos\left(1 \over n\right)\right] $
Note that $$ 0\le 1-\cos\frac{1}{n}=2\sin^2\frac{1}{2n}\le 2\cdot\left(\frac{1}{2n}\right)^2=\frac{1}{2n^2}. $$ We have used above that $$1-\cos (2x)=2\sin^2 x,$$ and also that $0 \le \sin x\le x$, whenever $x\in [0,\pi/2]$.
We can apply the Limit Test with $\rho =2$, $$\lim_{k \to \infty} k^2\left( 1 - \cos{\frac{1}{k}}\right) = \lim_{u \to 0}\frac{1 - \cos{u}}{u^2} = \frac{1}{2}$$ and thus the series converges absolutely.