Question

证ln(n+1)的n次方/n!<2/1+3/2+n+1/n

Answer 1

问：证ln(n+1)的n次方/n!<2/1+3/2+n+1/n

答：、令f(x) = ln(1+x) - x ，x>0f'(x) = 1/(1+x) -1<0故f(x)是减函数f(x) < f(0) =0即 ln(1+x) < x分别令x = 1，1/2，1/3，……得ln2 <1ln 3/2 < 1/2ln 4/3 < 1/3……ln (n+1)/n < 1/n各式相加得 ln (n+1) < 1+1/2+1/3+……+1/n2、令g(x)= lnx - (1- 1/x) ，x>1g'(x) = 1/x - 1/x² = 1/x (1 - 1/x) >0g(x)是增函数∴g(x) > g(1) = 0即 lnx > 1 - 1/x分别令x =2，3/2，4/3，……，n/(n-1)ln2 > 1/2ln(3/2) > 1/3ln(4/3) > 1/4……ln[n/(n-1)] > 1/n各式相加得 lnn > 1/2+1/3+....+1/n即 1+1/2+1/3+....+1/n < 1+ lnn故 ln(1+n) ＜1+1/2+1/3+....+1/n ＜ 1+ lnn