求证:1+1/2+1/3+...+1/n>In(n+1)+n/2(n+1) (n属于N+)

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求证:1+1/2+1/3+...+1/n>In(n+1)+n/2(n+1)(n属于N+)求证:1+1/2+1/3+...+1/n>In(n+1)+n/2(n+1)(n属于N+)求证:1+1/2+1/3

求证:1+1/2+1/3+...+1/n>In(n+1)+n/2(n+1) (n属于N+)
求证:1+1/2+1/3+...+1/n>In(n+1)+n/2(n+1) (n属于N+)

求证:1+1/2+1/3+...+1/n>In(n+1)+n/2(n+1) (n属于N+)
推荐:采用构造函数法证明.注意到ln(n+1)=ln[(n+1)/n]+ln[n/(n-1)]+...+ln(3/2)+ln(2/1),而n/(n+1)=1-1/(n+1)=(1-1/2)+(1/2-1/3)+...+[1/n-1/(n+1)].于是我们根据要证明的表达式,两边取通项(x-->1/n)构造函数f(x)=x-ln(1+x)-(1/2)[x-x/(x+1)],x>0,求导易得f'(x)=x^2/[2(x+1)^2]>0,x>0.于是f(x)在x>0上单调递增,又f(x)可在x=0处连续,则f(x)>f(0)=0,x>0得x-ln(1+x)-(1/2)[x-x/(x+1)]>0即x>ln(1+x)+(1/2)[x-x/(x+1)],x>0.再取1/n(>0)替换x有1/n>ln[(n+1)/n]+(1/2)[1/n-1/(n+1)],将此不等式从1到n项累加得1+1/2+1/3+...+1/n>{ln[(n+1)/n]+ln[n/(n-1)]+...+ln(3/2)+ln(2/1)}+(1/2){(1-1/2)+(1/2-1/3)+...+[1/n-1/(n+1)]}=ln(n+1)+(1/2)[1-1/(n+1)]=ln(n+1)+n/[2(n+1)],(n属于N+)命题便得证.觉得行可采纳.