若n是不小于2的正整数,试证4/7＜1-1/2+1/3-1/4+···+1/（2n-1）-1/2n＜√2/2 柯西不等式

f(n)=1-1/2+1/3-1/4+...+1/(2n-1)-1/2n
=(1+1/2+1/3+...+1/2n)-2(1/2+1/4+...+1/2n)
=(1+1/2+1/3+...+1/2n)-(1+1/2+...+1/n)
=1/(n+1)+1/(n+2)+...+1/2n
证明f(n)＞4/7
f(n)*[(n+1)+(n+2)+(n+3)+……+(2n)]≥(1+1+1+……+1)²=n²
f(n)*{[n*(3n+1)]/2}≥n²
f(n)≥2n²/[n*(3n+1)]=2n/(3n+1)＞2*2/(3*2+1)=4/7
证明f(n)＜√2/2
f²(n)={[1/(n+1)]*1+[1/(n+2)]*1+...+[1/2n]*1}²
≤(1²+1²+1²+……+1²)*[1/(n+1)²+1/(n+2)²+...+1/(2n)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-1)(2n)]
=n*[1/n-1/(n+1)+1/(n+1)-1/(n+2)+……+1/(2n-1)-1/2n]
=n*(1/n-1/2n)=n*1/2n=1/2
所以f²(n)＜1/2
所以f(n)＜√2/2
得证……
不明白的话,HI我……

f(n)=1-1/2+1/3-1/4+...+1/(2n-1)-1/2n
=(1+1/2+1/3+...+1/2n)-2(1/2+1/4+...+1/2n)
=(1+1/2+1/3+...+1/2n)-(1+1/2+...+1/n)
=1/(n+1)+1/(n+2)+...+1/2n
证明f(n)＞4/7
f(n)*[(n+1)+(n+2)+(n+3)+……...