数列问题 高人进

a(n+1)=((2n+1)/(2n+2))an=an-(1/(2(n+1))an
(n+1)a(n+1)-(n+1)an=-(1/2)an

(n+1)a(n+1)-nan-an=-(1/2)an
nan-(n-1)a(n-1)-a(n-1)=-(1/2)a(n-1)
(n-1)a(n-1)-(n-2)a(n-2)-a(n-2)=-(1/2)a(n-2)
....
2a2-2a1=-(1/2)a1
以上各式相加得:
(n+1)a(n+1)-(an+a(n-1)+...+a1)-a1=-(1/2)(an+a(n-1)+...+a1)
(n+1)a(n+1)-An-a1=-(1/2)An
An=(2/3)(a1-(n+1)a(n+1))=(2/3)((1/2)-(n+1)a(n+1))

而其中:(n+1)a(n+1)=(n+1)*(1*3*5...*(2n+1))/(2*4*6...*(2n+2))
=(2n+1)!/(2^(2n+1)*(n!)^2)

An=(1/3)-(1/3)((2n+1)!/(2^(2n)*(n!)^2))

有限的还是无限的?

见过这题的 你等下啊 我给你查到的