Ủa đề bài như này là sao bạn? Cho dãy x(k), nhưng lại đi tìm u(n)?
Ok start
\(\dfrac{1}{2!}=\dfrac{2!-1}{2!}=1-\dfrac{1}{2!};\dfrac{2}{3!}=\dfrac{1}{3}=\dfrac{3!-2!}{3!.2!}=\dfrac{1}{2!}-\dfrac{1}{3!}\)
\(\Rightarrow\dfrac{k}{\left(k+1\right)!}=\dfrac{1}{k!}-\dfrac{1}{\left(k+1\right)!}\)
Explain: \(\dfrac{1}{k!}-\dfrac{1}{\left(k+1\right)!}=\dfrac{\left(k+1\right)k!-k!}{k!\left(k+1\right)!}=\dfrac{k+1-1}{\left(k+1\right)!}=\dfrac{k}{\left(k+1\right)!}\)< Có nên xài quy nạp mạnh cho chặt chẽ hơn ko nhỉ?>
Nhớ lại 1 bài toán lớp 6 cũng có dạng như này
\(\Rightarrow x_k=1-\dfrac{1}{\left(k+1\right)!}\)
Xet \(x_{k+1}-x_k=1-\dfrac{1}{\left(k+2\right)!}-1+\dfrac{1}{\left(k+1\right)!}=\dfrac{1}{\left(k+1\right)!}-\dfrac{1}{\left(k+2\right)!}>0\Rightarrow x_{k+1}>x_k\)
\(\Rightarrow x_1< x_2< ...< x_{2011}\Rightarrow x_1^n< x_2^n< ...< x_{2011}^n\)
\(\Rightarrow\sqrt[n]{x_1^n+x_2^n+...+x_{2011}^n}< \sqrt[n]{x_{2011}^n+x^n_{2011}+...+x^n_{2011}}=\sqrt[n]{2011.x^n_{2011}}=x_{2011}.\sqrt[n]{2011}\)
Mat khac: \(x_{2011}=\sqrt[n]{x^n_{2011}}< \sqrt[n]{x_1^n+x_2^n+...+x_{2011}^n}\)
\(\Rightarrow x_{2011}< \sqrt[n]{x^n_1+x_2^n+...+x_{2011}^n}< \sqrt[n]{2011}x_{2011}\)
\(\lim\limits x_{2011}=1-\dfrac{1}{2012!}\)
\(\lim\limits\sqrt[n]{2011}x_{2011}=\lim\limits2011^0.x_{2011}=1-\dfrac{1}{2012!}\)
\(\Rightarrow\lim\limits\left(u_n\right)=1-\dfrac{1}{2012!}\)
Xin dung cuoc choi tai day, ban check lai xem dung ko, tinh tui hay au co khi sai :v