设 f(n) = 1/(n^2+n+1)+2/(n^2+n+2)+......+n/(n^2+n+n)),
f(n) >= (1+2+3+......+n) / (n^2+n+n) = n*(n+1) /2 / (n^2+n+n) = u(n)
f(n) <= (1+2+3+......+n) / (n^2+n+1) = n*(n+1) /2 / (n^2+n+1) = v(n)
Limit [ u(n), n ->∞] = Limit [ v(n), n ->∞] = 1/2
根据夹挤准则, Limit [ f(n), n ->∞] = 1/2 为所求。
为了方便叙述,原式我就简写为limf(n)了
显然f(n)>(1+2+……n)/(n^2+n+n)=(n^2+n)/(2*(n^2+n+n)………………A
又f(n)<(1+2+……n)/(n^2+n+1)=(n^2+n)/(2*(n^2+n+1)………………B
现在分别求A,B两式极限就很好求了
A,B两式求法相同,均对分子分母除n^2,易得limA=1/2,limB=1/2
即1/2