※ 引述《Intercome (今天的我小帅)》之铭言： : ___ __ __ : 若n属於正整数，利用递回数列关系求证√1+√2+√3+...+√n < 2 Let a_1=2, and a_(k+1)=a_k^2-k for k=1,2,... 依次去根号并移项可知, 原不等式等价於证明 a_n > 0 for all n>=1. 将证更强的命题: a_n > n for n>=1. pf. Hence a_2=3>2, a_3=9-2=7 > 3. Now, Suppose the statement is ture for k<=n (n>=3) Then a_(n+1) = a_n^2 - n > n^2-n = n+1 + (n-1)^2 -2 > n+1. (if n>=3) --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 131.215.6.212