已知数列an,满足a1=1,an=an-1+1/an-1,求证;对任意,不等式根号2n-1

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已知数列an,满足a1=1,an=an-1+1/an-1,求证;对任意,不等式根号2n-1
已知数列an,满足a1=1,an=an-1+1/an-1,求证;对任意,不等式根号2n-1

已知数列an,满足a1=1,an=an-1+1/an-1,求证;对任意,不等式根号2n-1
(1)a1+3*a2+5*a3+.+(2n-1)*an=pn
令n=1 a1=p,p=1
a1+3*a2+5*a3+.+(2n-1)*an=n
则a1+3*a2+5*a3+.+(2n-3)*an-1=n-1 (n≥2)
两式相减 (2n-1)*an=1
an=1/(2n-1) (n≥2) n=1时也成立
∴an=1/(2n-1)
(2) bn=an*an-1=1/(2n-1)(2n-3)=1/2[1/(2n-3)-1/(2n-1)] (n≥2)
sn=1/2[{(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+……+[1/(2n-3)-1/(2n-1)]}
=1/2[1-1/(2n-1)]
=(n-1)/(2n-1) (n>=2)