求证通项为根号(N平方+N)分之一的和的极限等于1,N趋向无穷

N/(N^2 + 1)^(1/2) > 1/(N^2 + 1)^(1/2) + 1/(N^2 + 2)^(1/2) + ...+ 1/(N^2 + N)^(1/2) > N/(N^2 + N)^(1/2),
lim_{N->+无穷}[N/(N^2 + 1)^(1/2)] = lim_{N->+无穷}[1/(N^(-2) + 1)^(1/2)] = 1
lim_{N->+无穷}[N/(N^2 + N)^(1/2)] = lim_{N->+无穷}[1/(N^(-1) + 1)^(1/2)] = 1
所以,
当N->+无穷时,1/(N^2 + 1)^(1/2) + 1/(N^2 + 2)^(1/2) + ...+ 1/(N^2 + N)^(1/2)的极限存在,且
lim_{N->+无穷}[1/(N^2 + 1)^(1/2) + 1/(N^2 + 2)^(1/2) + ...+ 1/(N^2 + N)^(1/2)] = 1

因为1/√(N^2+1)+1/√(N^2+2)+...+1/√(N^2+N
< 1/√N^2+1/√N^2+`...+1/√N^2 = 1
1/√(N^2+1)+1/√(N^2+2)+....+1/√(N^2+N)
> 1/√(N^2+N)+1/√(N^2+N)+...+1/√(N^2+N)
= N/(√N^2+N)
N无穷时，limN/(√N^2+...

1/√(N^2+1)+1/√(N^2+2)+`````+1/√(N^2+N
< 1/√N^2+1/√N^2+`````+1/√N^2 = 1
1/√(N^2+1)+1/√(N^2+2)+`````+1/√(N^2+N)
> 1/√(N^2+N)+1/√(N^2+N)+`````+1/√(N^2+N)
= N/√{(N+0.5)^2-0.25}
N无穷时，N/(N+0.5)=1
所以极限为1

我记得这个极限好像不等于1。