Question

n≥2 ，n为正整数时，求证4/7≤1/（n+1)+1/(n+2)+...+1/(2n-1)+1/2n＜√2/2

Answer 1

由[1/（n+1)+1/(n+2)+...+1/(2n-1)+1/(2n)][（n+1)+(n+2)+...+(2n)]
>[(n+1)/（n+1)+(n+2)/(n+2)+...+(2n)/(2n)]^2=n^2，
故1/（n+1)+1/(n+2)+...+1/(2n)>n^2/[（n+1)+(n+2)+...+(2n)]=n^2/[n(3n+1)/2]
=2/(3+1/n)≥2/(3+1/2)=4/7

又[1/（n+1） + 1/（n+2）+...+ 1/2n]^2
={[1/（n+1）]*1 + [1/（n+2）]*1+...+ [1/2n]*1}^2
<(1^2*n)*[1/（n+1）^2 + 1/（n+2）^2+...+ 1/(2n)^2]
<n*[1/n(n+1)+1/(n+1)(n+2)+...+1/2n(2n-1)]
=n*[1/n-1/(n+1)+1/(n+1)-1/(n+2)+...+1/(2n-1)-1/2n)]
=n*(1/n-1/2n)
=1/2
∴1/（n+1)+1/(n+2)+...+1/(2n-1)+1/2n＜√2/2