Đặt f(n)=(n^2+n+1)^2+1 . Xét dãy số (un) sao cho u(n)=f(1).f(3).f(5)...f(2n-1)/f(2).f(4).f(6)...f(2n) Tính limn căn bậc hai u(n).

* Hướng dẫn giải

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

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

$\Rightarrow u_n=\frac{2}{10}.\frac{10}{26}.\frac{26}{50}\mathrm{....}\frac{{\left(2n-3\right)}^2+1}{{\left(2n-1\right)}^2+1}.\frac{{\left(2n-1\right)}^2+1}{{\left(2n+1\right)}^2+1}=\frac{2}{{\left(2n+1\right)}^2+1}$

$\Rightarrow \lim n\sqrt{u_n}=\lim \sqrt{\frac{2n^2}{4n^2+4n+2}}=\frac{1}{\sqrt{2}}.$