序列 $\sed{x_n}$ 是正项单调递增并且有界, 证明级数 $\dps{\vsm{n}\sex{1-\frac{x_n}{x_{n+1}}}}$ 收敛. (国外赛题)
证明: 由 $\dps{1-\frac{x_n}{x_{n+1}}\leq 0}$ 及 $$\beex \bea \sum_{k=1}^n \sex{1-\frac{x_k}{x_{k+1}}} &\leq \sum_{k=1}^n \frac{x_{k+1}-x_k}{x_{k+1}} \leq \frac{1}{x_k}\sum_{k=1}^n (x_{k+1}-x_k) =\frac{1}{x_1}(x_{n+1}-x_1)\\ &\leq \frac{1}{x_1}\sex{\sup_{n\geq 1}x_n-x_1} \eea \eeex$$ 即知结论成立.
