Đặt

D

- Hướng dẫn giải

Đáp án D.

Ta có phân tích $f\left(n\right)={\left(n^2+n+1\right)}^2+1=n^4+2n^2\left(n+1\right)+{\left(n+1\right)}^2+1$ 

$=n^2\left(n^2+2n+2\right)+{\left(n+1\right)}^2+1=n^2\left[{\left(n+1\right)}^2+1\right]+{\left(n+1\right)}^2+1=\left(n^2+1\right)\left[{\left(n+1\right)}^2+1\right]$ 

Khi đó $\frac{f\left(2k-1\right)}{f\left(2k\right)}=\frac{{\left(2k-1\right)}^2+1}{{\left(2k+1\right)}^2+1}\Rightarrow u_n=\frac{1^2+1}{3^2+1}.\frac{3^2+1}{5^2+1}...\frac{{\left(2n-1\right)}^2+1}{{\left(2n+1\right)}^2+1}=\frac{1}{2n^2+2n+1}$ 

$\Rightarrow lim\left(n\sqrt{u_n}\right)=lim\frac{n}{\sqrt{2n^2+2n+1}}=\frac{1}{\sqrt{2}}.$