作业帮数列{根号（ n+2）-2根号（n+1）+根号n},求前n项和的极限
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数列{根号（ n+2）-2根号（n+1）+根号n},求前n项和的极限
a(n) = [(n+2)^(1/2) - (n+1)^(1/2)] - [(n+1)^(1/2) - n^(1/2)],
s(n) = a(1)+a(2)+...+a(n-1)+a(n)
=[3^(1/2)-2^(1/2)]-[2^(1/2)-1^(1/2)] + [4^(1/2)-3^(1/2)]-[3^(1/2)-2^(1/2)] + ...+[(n+1)^(1/2)-n^(1/2)]-[n^(1/2)-(n-1)^(1/2)] + [(n+2)^(1/2)-(n+1)^(1/2)]-[(n+1)^(1/2)-n^(1/2)]
=[(n+2)^(1/2)-(n+1)^(1/2)] - [2^(1/2)-1^(1/2)],
=1/[(n+2)^(1/2) + (n+1)^(1/2)] - 2^(1/2) + 1,
n->无穷大时,1/[(n+2)^(1/2)+(n+1)^(1/2)] -> 0,
所以,n->无穷大时,
s(n) -> 0 - 2^(1/2) + 1 = 1-2^(1/2)
s(n) = a(1)+a(2)+...+a(n-1)+a(n)
=[3^(1/2)-2^(1/2)]-[2^(1/2)-1^(1/2)] + [4^(1/2)-3^(1/2)]-[3^(1/2)-2^(1/2)] + ...+[(n+1)^(1/2)-n^(1/2)]-[n^(1/2)-(n-1)^(1/2)] + [(n+2)^(1/2)-(n+1)^(1/2)]-[(n+1)^(1/2)-n^(1/2)]
=[(n+2)^(1/2)-(n+1)^(1/2)] - [2^(1/2)-1^(1/2)],
=1/[(n+2)^(1/2) + (n+1)^(1/2)] - 2^(1/2) + 1,
n->无穷大时,1/[(n+2)^(1/2)+(n+1)^(1/2)] -> 0,
所以,n->无穷大时,
s(n) -> 0 - 2^(1/2) + 1 = 1-2^(1/2)
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