Phương pháp giải:

- Biến đổi, rút gọn biểu thức\({{v}_{n}}\) rồi tính giới hạn.

Lời giải chi tiết:

\({{u}_{2}}=\sqrt{1.2.3.4+1}=5,\,\,{{u}_{n}}>0,\,\,\forall n=1;2;...\)

Ta có:

\(\begin{align}  & {{u}_{n+1}}=\sqrt{{{u}_{n}}\left( {{u}_{n}}+1 \right)\left( {{u}_{n}}+2 \right)\left( {{u}_{n}}+3 \right)+1}=\sqrt{\left( u_{n}^{2}+3{{u}_{n}} \right)\left( u_{n}^{2}+3{{u}_{n}}+2 \right)+1} \\ & =\sqrt{{{\left( u_{n}^{2}+3{{u}_{n}} \right)}^{2}}+2\left( u_{n}^{2}+3{{u}_{n}} \right)+1}=\sqrt{{{\left( u_{n}^{2}+3{{u}_{n}}+1 \right)}^{2}}}=u_{n}^{2}+3{{u}_{n}}+1 \\ & \Rightarrow {{u}_{n+1}}+1=u_{n}^{2}+3{{u}_{n}}+2=\left( {{u}_{n}}+1 \right)\left( {{u}_{n}}+2 \right) \\ & \Rightarrow \frac{1}{{{u}_{n+1}}+1}=\frac{1}{\left( {{u}_{n}}+1 \right)\left( {{u}_{n}}+2 \right)}=\frac{1}{{{u}_{n}}+1}-\frac{1}{{{u}_{n}}+2} \\ & \Rightarrow \frac{1}{{{u}_{n}}+2}=\frac{1}{{{u}_{n}}+1}-\frac{1}{{{u}_{n+1}}+1} \\\end{align}\)

Do đó: \({{v}_{n}}=\sum\limits_{i=1}^{n}{\frac{1}{{{u}_{i}}+2}=}\sum\limits_{i=1}^{n}{\left( \frac{1}{{{u}_{i}}+1}-\frac{1}{{{u}_{i+1}}+1} \right)=\frac{1}{{{u}_{1}}+1}-\frac{1}{{{u}_{n+1}}+1}=\frac{1}{2}-\frac{1}{{{u}_{n+1}}+1}}\)

Xét hiệu \({{u}_{n+1}}-{{u}_{n}}=u_{n}^{2}+3{{u}_{n}}+1-{{u}_{n}}={{\left( {{u}_{n}}+1 \right)}^{2}}>0\Rightarrow \left( {{u}_{n}} \right)\) là dãy tăng.

Giả sử \(\lim {{u}_{n+1}}=\lim {{u}_{n}}=a>0\Rightarrow a={{a}^{2}}+3a+1\Rightarrow {{a}^{2}}+2a+1=0\Leftrightarrow a=-1\,\,\left( ktm \right)\Rightarrow \lim {{u}_{n}}=+\infty \) 

\(\Rightarrow \lim {{v}_{n}}=\frac{1}{2}-\frac{1}{{{u}_{n+1}}+1}=\frac{1}{2}-0=\frac{1}{2}.\)

Chọn C.