Re: 極限
※ 引述《HUJIKOLP (UUOJKLJLKJK)》之銘言:
: a0=0 a1=1 an=((an-1)+(an-2))/2
: 求liman
: n~~無窮
: 我這堤算2/3
: 不過沒答案@@!不知大家算多少
有個條件你沒打出來!!!!!!!!!!!!!!!!!!
n=2,3,4,......
題意知,
a_n-a_(n-1)=-(a_n-a_(n-2))
a_2-a_1 =-(a_2-a_0)
a_3-a_2=-(a_3-a_1)
a_4-a_3=-(a_4-a_2)
. .
. .
. .
a_(n-1)-a_(n-2)= -(a_(n-1)-a_(n-3))
+) a_n - a_(n-1) = - (a_n-a_(n-2))
------------------------------------------
a_n-a_1 = -a_n-a_(n-1)+a_0+a_1
=>2a_n+a_(n-1)=2a_1+a_0=2
lim lim
因為 a_n = a_(n-1)
n->∞ n->∞
lim
所以 a_n= 2/3
n->∞ #
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