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不等式已知:n≥2，证明:4/7<1-1/2+1/3-1/

2012-09-17 21:51:07s***
已知:n≥2，证明:4/7<1-1/2+1/3-1/4+···+1/(2n-1)-1/2n<√2/2。不等式已知:n≥2，证明:4/71-1/2+1/3-1/4+···+1/(2n-1)-1/2n√2/2。：证明： `f(n)=1-1/2+1/?

最佳回答

  •   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。
      
    2012-09-17 22:17:35
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