1/[(n+1)n^1/2+n(n+1)^1/2]
=[(n+1)√n-n√(n+1)]/n(n+1)
=1/√n-1/√(n+1)
p=1/(2+2^1/2)+1/(18^1/2+12^1/2)+.+1/[(n+1)n^1/2+n(n+1)^1/2]
=(1-1/√2)+(1/√2-1/√3)+...+(1/√n-1/√(n+1))
=1-1/√(n+1)
所以,0