Đặt \(f\left( n \right) = {\left( {{n^2} + n + 1}...

D

- Hướng dẫn giải

Xét \(g\left( n \right) = \frac{{f\left( {2n - 1} \right)}}{{f\left( {2n} \right)}} \Rightarrow g\left( n \right) = \frac{{{{\left( {4{n^2} - 2n + 1} \right)}^2} + 1}}{{{{\left( {4{n^2} + 2n + 1} \right)}^2} + 1}}\)

Đặt  \(\left. \begin{array}{l}a = 4{n^2} + 1\\b = 2n\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}a \pm 2b = {\left( {2n \pm 1} \right)^2}\\a = {b^2} + 1\end{array} \right.\)

\( \Rightarrow g\left( n \right) = \frac{{{{\left( {a - b} \right)}^2} + 1}}{{{{\left( {a + b} \right)}^2} + 1}} = \frac{{{a^2} - 2ab + {b^2} + 1}}{{{a^2} + 2ab + {b^2} + 1}} = \frac{{{a^2} - 2ab + a}}{{{a^2} + 2ab + a}} = \frac{{a - 2b + 1}}{{a + 2b + 1}} = \frac{{{{\left( {2n - 1} \right)}^2} + 1}}{{{{\left( {2n + 1} \right)}^2} + 1}}\)

\( \Rightarrow {u_n} = \prod\limits_{i = 1}^n {g\left( i \right)}  = \frac{2}{{10}}.\frac{{10}}{{26}}...\frac{{{{\left( {2n - 1} \right)}^2} + 1}}{{{{\left( {2n + 1} \right)}^2} + 1}} = \frac{2}{{{{\left( {2n + 1} \right)}^2} + 1}}\)

\( \Rightarrow \lim n\sqrt {{u_n}}  = \lim \sqrt {\frac{{2{n^2}}}{{4{n^2} + 4n + 2}}}  = \frac{1}{{\sqrt 2 }}\)