[Toán 11]Dãy số.

R

rua_it

cho U1=1 ,U(n+1) = ( Un +2)/( Un + 1). xet tinh' tang giam~ va` bi chan
u2=32u_2=\frac{3}{2}

u3=32+232+1=75<32u_3=\frac{\frac{3}{2}+2}{\frac{3}{2}+1}=\frac{7}{5}<\frac{3}{2}

Dễ thấy dãy un u_n \exists cận trên và cận dưới:

1un32() 1 \leq u_n \leq \frac{3}{2} (*)

Với n=1,2 thì hiển nhiên () (*) đúng.

Giả sử () (*) đúng khi n=k(k2)n=k (k \geq 2)

1uk32\Rightarrow 1 \leq u_k \leq \frac{3}{2}

Cần chứng minh: 1uk+132(1) 1 \leq u_{k+1} \leq \frac{3}{2}(1)

uk+1=uk+2uk+1uk=2uk+1uk+11u_{k+1}=\frac{u_k+2}{u_k+1} \Rightarrow u_k=\frac{2-u_{k+1}}{u_{k+1}-1}

(1)12uk+1uk+1132\Rightarrow (1) \Leftrightarrow 1 \leq \frac{2-u_{k+1}}{u_{k+1}-1} \leq \frac{3}{2}

[tex]\Rightarrow \left{\begin{\frac{2-u_{k+1}}{u_{k+1}-1} \geq 1}\\{\frac{2-u_{k+1}}{u_{k+1}-1} \leq \frac{3}{2}}[/tex]

[tex]\Rightarrow \left{\begin{\frac{3-2u_{k+1}}{u_{k+1}-1} \geq 0}\\{\frac{7-5u_{k+1}}{u_{k+1}-1} \leq 0}[/tex]

[tex]\Rightarrow \left{\begin{\left[\begin{u_{k+1} <1}\\{u_{k+1} > \frac{7}{5}}}\\{1 \leq u_{k+1} \leq \frac{3}{2}}[/tex]

1un32\Rightarrow 1 \leq u_n \leq \frac{3}{2}

Xét hiệu: un+1un=un+2un+1unu_{n+1}-u_n=\frac{u_n+2}{u_n+1}-u_n

=un+2un.(un+1)un+1=un2+2un+1=\frac{u_n+2-u_n.(u_n+1)}{u_n+1}=\frac{-u_n^2+2}{u_n+1}

....

Gõ mấy cái tex chán:(
 
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