楼上的证明了
lim ∑i^p/n^(p+1)yn
yn->∞
(x(n+1)-xn)/(y(n+1)-yn)=
=n^p/[(n+1)^(p+1)-n^(p+1)]=
=n^p/[(p+1)n^p+...]
lim (x(n+1)-xn)/(y(n+1)-yn)= 1/(p+1)
故lim ∑i^p/n^(p+1)=lim xn/yn= lim (x(n+1)-xn)/(y(n+1)-yn)= 1/(p+1)