Re: [微積] 數列要如何證明收斂?

[Desperato (Farewell)]

※ 引述《shaqgod (ddinin)》之銘言：
: 請問各位大大3.4題要如何證明收斂
: http://m.imgur.com/gallery/RmLoD0X
: -----
: Sent from JPTT on my Asus ASUS_T00N.


3.

Let z_0 = y_0 = a
    z_1 = y_1 - y_0 = b-a
    z_n = y_n - y_(n-1), n positive integer

           -1
Then z_n = --- z_(n-1)
            3

            -1
         = (---)^(n-1) z_1   (z_0 is not in recursive formula)
             3
       n                   n   -1
y_n = sum z_k = z_0 + z_1 sum (---)^(k-1)
      k=0                 k=1   3

                   -1                          1        3
Since the series {(---)^(n-1)} converges to -------- = ---
                    3                       1-(-1/3)    4

the sequence {y_n} converges to  a + (3/4)(b-a) = (a+3b)/4



4.

Clearly x_n <= x_(n-1) for n >= 2
and x_n all positive thus x_n >= 0 for all n
By Monotone sequence theorem {x_n} converges to some real number


y_n >= y_(n-1) for n >= 2
We now try to find an upper bound M such that y_n <= M for all n

Let 1 >= k > 0
                              M
If there is an M such that ------- > 1 + k
                           y_(n-1)

      M      1 + k            2^n          2^n - 1
Then --- = --------- = (1+k)------- > (1+k)-------
     y_n   1 + 1/2^n        2^n + 1          2^n

                      1              1+k              2
         = (1+k)(1 - ---) = 1 + k - ----- >= 1 + k - ---
                     2^n             2^n             2^n

                 2     k          1
Now we want k - --- = ---, k = -------
                2^n    2       2^(n-2)

       M          1
Also ----- > 1 + --- = 2, take M = 4
      y_1        2^0


Then above is exactly the induction proof

      M            1
for ----- > 1 + ------- for all n
     y_n        2^(n-1)


Thus M > y_n for all n, M is an upper bound for {y_n}

By Monotone sequence theorem {y_n} converges to some real number

--

嗯嗯ow o

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推 shaqgod     : 謝謝
→ yyc2008     : M要怎麼找?
→ yyc2008     : 我知道了
推 cuttlefish  : yn*(1-1/2)=1-1/2^{n+1}
→ yyc2008     : c大的作法必須是y_n=Π(1+1/2^(2n))才行吧?
推 cuttlefish  : 喔我看錯了
→ yyc2008     : 沒關係,我一開始也看錯了
推 znmkhxrw    : 這題目給x_n跟y_n, x_n好證
→ znmkhxrw    : 把x_n跟y_n相乘 會得到Prod_{k=1~n} (1-1/4^k)
→ znmkhxrw    : 這個級數也是遞減的 得出y_n收斂
→ yyc2008     : 我原本也這樣想 但是{n},{1/n^2} 兩者相乘{1/n}收斂
→ yyc2008     : 但是{n}收斂?
推 znmkhxrw    : 你這例子有問題是因為1/n^2收斂到0
→ znmkhxrw    : 不過確實我要用那個方法 比需要確保x_n→L>0
→ znmkhxrw    : 這樣也是要先估出x_n的正下界QQ
→ yyc2008     : 嗯嗯 我的例子也有點問題