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=1a1+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=1an=1/(2n-1) (n≥2) n=1时也成立∴an=1/(2n-1)(2) bn=an*an-1=1/(2n-1)(2n...