Cho dãy \(({x_k})\) được xác định như sau: \({x_k} = \dfrac{1}{{2!}} + \dfrac{2}{{3!}} + ... + \dfrac{k}{{(k + 1)!}}\)

Tìm \(\lim {u_n}\) với \({u_n} = \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}}\).

Ta có: \(\dfrac{k}{{(k + 1)!}} = \dfrac{1}{{k!}} - \dfrac{1}{{(k + 1)!}}\) nên \({x_k} = 1 - \dfrac{1}{{(k + 1)!}}\)

$\begin{array}{l} \Rightarrow {x_k} - {x_{k + 1}} = 1 - \dfrac{1}{{\left( {k + 1} \right)!}} - 1 + \dfrac{1}{{(k + 2)!}} = \dfrac{1}{{(k + 2)!}} - \dfrac{1}{{(k + 1)!}} < 0\\ \Rightarrow {x_k} < {x_{k + 1}} \Rightarrow {x_1} < {x_2} < ... < {x_{2011}}\\ \Rightarrow x_1^n < x_2^n < ... < x_{2011}^n \Rightarrow \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}} < \sqrt[n]{{x_{2011}^n + x_{2011}^n + ... + x_{2011}^n}} = \sqrt[n]{{2011.x_{2011}^n}} = \sqrt[n]{{2011}}.{x_{2011}}\end{array}$

Lại có: \({x_{2011}} = \sqrt[n]{{x_{2011}^n}} < \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}}\)

Vậy: \({x_{2011}} < \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}} < \sqrt[n]{{2011}}{x_{2011}}\)

Ta có: \({x_{2011}} = 1 - \dfrac{1}{{\left( {2011 + 1} \right)!}} = 1 - \dfrac{1}{{2012!}}\)

\(\begin{array}{l} \Rightarrow \lim {x_{2011}} = {x_{2011}} = 1 - \dfrac{1}{{2012!}}\\ \Rightarrow \lim \sqrt[n]{{2011}}{x_{2011}} = \lim \,\,{2011^{\dfrac{1}{n}}}{x_{2011}} = {2011^0}.{x_{2011}} = {x_{2011}} = 1 - \dfrac{1}{{2012!}}\end{array}\)

Vậy \(\lim {u_n} = 1 - \dfrac{1}{{2012!}}\).