※ 引述《BIEGABOY (BIEGABOY)》之銘言：
: Let the sequence an=(1+1/n)^n
: 1.show that an is increasing.
Proof:
Let f(x)=(1+1/x)^x f(x_n)=a_n x_1=1,x_2=2...,x_n=n
取自然對數 F(x)=ln[f(x)]=x*ln(1+1/x)=x*ln[(x+1)/x ] (裡面變假分式)
遞增遞減性仍與原函數相同
x x-(x+1)
考慮 F'(x)=ln[(x+1)/x ]+x* -------------- * --------------
x + 1 x^2
=ln[(x+1)/x ] - (1/x+1)
claim : F'(x) > 0 for all x>=1
since 1. F'(1)=ln2-1/2 >0
2. F"(x)= (x/x+1) - (-1)*(x+1)^(-2) > 0
所以F'(x) >0 => F(x) 遞增 => f(x)遞增
=> f(x_n+1) > f(x_n)
=> a_n+1 > a_n for all n>=1
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.113.93.114
※ 編輯: gkaok2 來自: 140.113.22.70 (02/09 08:53)