摘要
高中《代数》第二册112页11题是:证明1+1/(2<sup>1/2</sup>)+1/(3<sup>1/2</sup>)+…+1/(n<sup>1/2</sup>)】n<sup>1/2</sup>,（n】1）.文[1]给出了比上式更强的结论:2(（n+1）<sup>1/2</sup>-1)【sum from k=1 to n 1/(k<sup>1/2</sup>)【2n<sup>1/2</sup>-1,（n】1）。（Ⅰ） 本文对（Ⅰ）式进行加强,从而把（Ⅰ）式的结论统一到本文结论之中。且给出估计和式sum from k=1 to n 1/(K<sup>1/2</sup>)值（绝对误差不超过0.16）的一种方法。由1°,2°知（Ⅱ）式成立。 （Ⅱ）式亦可用数学归纳法证明。容易证明 (（n+1）<sup>1/2</sup>)+n<sup>1/2</sup>-2<sup>1/2</sup>【2(n<sup>1/2</sup>)-1,(（n+2）<sup>1/2</sup>)+n<sup>1/2</sup>-3<sup>1/2</sup>】2(（n+1）<sup>1/2</sup>-1).所以,（Ⅰ）式可看成是（Ⅱ）式的直接推论。因为 0【(（n+1）<sup>1/2</sup>)+n<sup>1/2</sup>-2<sup>1/2</sup>) -((（n+2）<sup>1/2</sup>)+n<sup>1/2</sup>-3<sup>1/2</sup>) =(（n+1）<sup>1/2</sup>-（n+2）<sup>1/2</sup>+(3<sup>1/2</sup>-2<sup>1/2</sup>) 【3<sup>1/2</sup>-2<sup>1/2</sup>【0.32。所以用 [(（n+1）<sup>1/2</sup>+n<sup>1/2</sup>-2<sup>1/2</sup>)+(（n+2）<sup>1/2</sup>+n<sup>1/2</sup>
