不等式已知:n≥2,证明:4/7
证明:
`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
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
f&...全部
证明:
`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
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
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
∴f²(n)<1/2
∴f(n)<√2/2
得证。
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