Re: [高微] 證明數列收斂
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 = Sn + a_n+1
=> Sn is increasing and bounded above
=> lim Sn = M exists
n->∞
for all ε> 0 , there is a k€N ，s.t. |Sn-M|<ε/2 for all n≧k
|S_n+1 - Sn|=|S_n+1 -M +M -Sn|≦|S_n+1 -M|+|M -Sn|≦ε/2 + ε/2 = ε
=> lim ( S_n+1 - Sn ) = 0
n->∞
感覺這樣寫就可以了..不知道這個方向的想法是否正確?
※ 引述《IminXD (Encore LaLa)》之銘言：
: 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
: 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
: ==> <a_n> is a increasing seq.
: Sn + Sn ≦ S_n-1 + S_n+1
: => Sn - S_n+1 ≦ S_n-1 - Sn
: Since Sn be a bounded sequence， |Sn| ≦ M with M€R ((屬於不會打QQ
: => |a_n+1| ≦ |Sn - S_n+1| ≦ |Sn| + |S_n+1| ≦ 2M
: => <a_n> is bounded
: Cause <a_n> is increasing and bounded above
: => <a_n> converges
: 後面的部份我不知道該怎麼證明lim (S_n+1 - Sn) = 0
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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.42.91.161
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11/29 21:33, , 1F
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所以我在用Since S_n+1 = Sn + a_n+1就該先討論嗎?
if a_n ≧ 0，S_n+1 = Sn + a_n+1 for all n€N
=> Sn is increasing and bounded above
if a_n < 0，S_n+1 = Sn + a_n+1 for all n€N
=> Sn is decreasing and bounded below
and so lim Sn = M exists
如果這樣子寫呢? 因為我從式子裡面看不出來a_n是恆負這件事阿QQ
※ 編輯: IminXD 來自: 114.42.91.161 (11/29 22:33)
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11/29 23:19, , 7F
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11/29 23:47, , 9F
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