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討論串[高微] 證明數列收斂
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2S_n<=S_n-1+S_n+1 可推得S_n-S_n-1<=S_n+1-S_n. a_n<=a_n+1 a_n為非嚴格遞增數列。. 故聯想到least upper bound property. claim a_n bounded. 因s_n bounded. a_n=S_n-S_n-1 |a_
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Proof. Let a_n ＝ s_(n+1) - s_n, it is easy to see that {a_n} converges since. it is increasing and bounded. The limit of {a_n} is denoted by a. We. wi
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Let A=lim( s(n+1) - s(n) ), which exists by the assumption,. and suppose |s(i)|<B for all i.. Let m be any natural number, then. m-1. s(n+m)-s(n)=sum
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今天想了一下，做a_n會卡，所以乾脆用Sn來做試試看. 上半部一樣. : If 2Sn ≦ S_n-1 + S_n+1. : => Sn + Sn ≦ S_n-1 + S_n+1. : => Sn - S_n-1 ≦ S_n+1 - Sn. : => a_n ≦ a_n+1. Since S_n+1
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題目:Let Sn be a bounded sequence of real numbers. Assume 2Sn≦S_n-1 + S_n+1. Show that lim ( S_n+1 - Sn ) = 0. n->∞. 我的作法是這樣... Let a_n+1 = Sn+1 - S_n.
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