设n为自然数,求证n+1分之1+n+2分之1+n+3分之1+...+3n分之1大于4n+1分之4n

第一项与倒数第一项相加,第二项与倒数第二项相加.[1/n+1]+[1/n+2]+.+[1/3n]=[1/n+1]+[1/3n] +[1/n+2]+[1/3n-1] +.+[1/2n]+[1/2n+1]=(4n+1){1/(n+1)(3n) +1/(n+2)(3n-1) +.+1/(2n)(2n+1) }>(4n+1)*n*1/(2n)(2n+1)=(4n...