Ta sẽ chứng minh \(u_n>1\)
- Với \(n=1\Rightarrow u_1=2>1\) đúng
- Giả sử \(u_k>1\)
- Ta cần chứng minh \(u_{k+1}=\frac{u_k^2}{2u_k-1}>1\)
Thật vậy, \(u_{k+1}-1=\frac{u_k^2}{2u_k-1}-1=\frac{\left(u_k-1\right)^2}{2u_k-1}>0\) \(\Rightarrow u_{k+1}>1\)
Vậy dãy bị chặn dưới bởi 1
\(u_{n+1}-u_n=\frac{u_n^2}{2u_n-1}-u_n=\frac{u_n-u_n^2}{2u_n-1}=\frac{u_n\left(1-u_n\right)}{2u_n-1}< 0\) do \(u_n>1\)
\(\Rightarrow u_{n+1}< u_n\Rightarrow\) dãy giảm
Dãy giảm và bị chặn dưới nên dãy bị chặn
b/
Ta có: \(u_{n+1}-1=\frac{\left(u_n-1\right)^2}{2\left(u_n-1\right)+1}\)
\(\frac{1}{u_{n+1}-1}=\frac{2\left(u_n-1\right)+1}{\left(u_n-1\right)^2}\Leftrightarrow\frac{1}{u_{n+1}-1}+1=\frac{2\left(u_n-1\right)+1}{\left(u_n-1\right)^2}+1=\frac{\left(u_n-1\right)+2\left(u_n-1\right)+1}{\left(u_n-1\right)^2}\)
\(\Leftrightarrow\frac{u_{n+1}}{u_{n+1}-1}=\left(\frac{u_n}{u_n-1}\right)^2\)
Đặt \(v_n=\frac{u_n}{u_n-1}\Rightarrow\left\{{}\begin{matrix}v_1=2\\v_{n+1}=v_n^2\end{matrix}\right.\)
\(v_n=v_{n-1}^{2^1}=v_{n-2}^{2^2}=v_{n-3}^{2^3}=...=v_1^{2^{n-1}}=2^{2^{n-1}}\)
\(\Rightarrow\frac{u_n}{u_n-1}=2^{2^{n-1}}\Rightarrow u_n=\frac{2^{2^{n-1}}}{2^{2^{n-1}}-1}\)
