数列题求解答！！数学！！！ 10
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1.
an=1/n
bn=S(2n)-Sn
b(n+1)-bn=S(2n+2)-S(n+1)-[S(2n)-Sn]
=a(n+2)+a(n+3)+...+a(2n+2)-[a(n+1)+a(n+2)+...+a(2n)]
=a(2n+1)+a(2n+2)-a(n+1)
=1/(2n+1)+ 1/(2n+2) -1/(n+1)
=1/(2n+1)+ 1/(2n+2) -2/(2n+2)
=1/(2n+1) -1/(2n+2)
=1/[(2n+1)(2n+2)]
>0
b(n+1)>bn
数列{bn}单调递增
an=1/n
bn=S(2n)-Sn
b(n+1)-bn=S(2n+2)-S(n+1)-[S(2n)-Sn]
=a(n+2)+a(n+3)+...+a(2n+2)-[a(n+1)+a(n+2)+...+a(2n)]
=a(2n+1)+a(2n+2)-a(n+1)
=1/(2n+1)+ 1/(2n+2) -1/(n+1)
=1/(2n+1)+ 1/(2n+2) -2/(2n+2)
=1/(2n+1) -1/(2n+2)
=1/[(2n+1)(2n+2)]
>0
b(n+1)>bn
数列{bn}单调递增
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