Re: [分析] 初微(28)
※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: Suppose that |a_n| < 2 and
: |a_(n+2) - a_(n+1)|≦ (1/8)| ( a_(n+1) )^2 - ( a_n )^2 |
: for all positive integers n. Show that {a_n} converges.
Hence |a_1-a_0|<= 1/2
we prove by induction that |a_k-a_(k-1)| <= 1/2^k
as following:
|a_k-a_(k-1)| <= 1/8|a_(k-1)-a_(k-2)||a_(k-1)+a_(k-2)|
<= 1/2^(k+2)*=1/2^k.
and therefore, |a_i-a_j| <= 1/2^i+...+2^(j-1) <= 1/2^(i-1). QED.
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