设 $$ex x_n=sum_{k=2}^n frac{cos k}{k(k-1)}, eex$$ 判断 $sed{x_n}$ 是否收敛?
解答: 由 $$eex ea |x_{n+p}-x_n|&=sev{sum_{k=n+1}^{n+p}frac{cos k}{k(k-1)}} leq sum_{k=n+1}^{n+p}frac{1}{k(k-1)} =sum_{k=n+1}^{n+p}sex{frac{1}{k-1}-frac{1}{k}} =frac{1}{n}-frac{1}{n+p}\ &<frac{1}{n} o 0quadsex{n oinfty} eea eeex$$ 及 Cauchy 收敛准则即知 $sed{x_n}$ 收敛.