We use the fact that \(\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}. \)

By Heine's theorem, \(\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}\text{ and } \lim_{n\to\infty}\frac{1}{n}=0 \text{ imply that } \lim_{n\to\infty}\frac{1-\cos\tfrac{1}{n}}{(\tfrac{1}{n})^2}=\frac{1}{2}. \)

So if we let \(\displaystyle a_n=\left(1-\cos\left(\frac{1}{n}\right)\right)n^{\alpha}\), \(\displaystyle b_n=\frac{1}{n^2}n^{\alpha}, \) then we easily obtain \(\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{1-\cos\tfrac{1}{n}}{(\tfrac{1}{n})^2} =\frac{1}{2}\in(0,\infty). \)

By the limit comparison test, the given series converges if and only if the series \(\displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}n^{\alpha-2} \) converges. This series will converge when \(\alpha-2<-1\) or \(\alpha<1\).