根号下n的平方加n分之n的从0到无穷的极限为什么等于1
2015-10-31
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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
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
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