难题 数列 极限：证明若p为自然数,则 lim (∑i^p/n^p)-n/(p+1)=1/2

(∑i^p/n^p)-n/(p+1)=
=[(p+1)(1^p+2^p+...+n^p)-n^(p+1)]/[(p+1)*n^p]
设xn=(p+1)(1^p+2^p+...+n^p)-n^(p+1)
yn=(p+1)*n^p
没错,要用到lim xn/yn=lim (x(n+1)-xn)/(y(n+1)-yn)
(x(n+1)-xn)/(y(n+1)-yn):
x(n+1)-xn=[(p+1)(1^p+2^p+...+(n+1)^p)-(n+1)^(p+1)]-[(p+1)(1^p+2^p+...+n^p)-n^(p+1)]=
=(p+1)(n+1)^p+n^(p+1)-(n+1)^(p+1)
(n+1)^(p+1)=n^(p+1)+(p+1)n^(p)+p(p+1)/2*x^(p-1)+...
x(n+1)-xn=-p(p+1)/2*n^(p-1) -...
y(n+1)-yn=-p(p+1)*n^(p-1) -...
lim (x(n+1)-xn)/(y(n+1)-yn)=-p(p+1)/2*n^(p-1) /-p(p+1)*n^(p-1)=1/2
故
lim xn/yn=lim (x(n+1)-xn)/(y(n+1)-yn)=1/2