Phương pháp giải:

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

Lời giải chi tiết:

Xét \({u_{n + 1}} - {u_n} = \frac{{\sqrt {u_n^2 + 4{u_n}}  + {u_n}}}{2} - {u_n} = \frac{{\sqrt {u_n^2 + 4{u_n}}  - {u_n}}}{2} > \frac{{\sqrt {u_n^2}  - {u_n}}}{2} = 0 \Rightarrow \left( {{u_n}} \right)\) là dãy tăng.

Giả sử \(\lim {u_n} = a\) thì \(a > 0\) và \(a = \frac{{\sqrt {{a^2} + 4a}  + a}}{2} \Leftrightarrow a = \sqrt {{a^2} + 4a} \)\( \Leftrightarrow {a^2} = {a^2} + 4a \Rightarrow a = 0\) (vô lý).

Suy ra \(\lim {u_n} =  + \infty \)

\(\begin{array}{l}\,\,\,\,\,\,\,{u_n} = \frac{{\sqrt {u_{n - 1}^2 + 4{u_{n - 1}}}  + {u_{n - 1}}}}{2}\\ \Leftrightarrow 2{u_n} - {u_{n - 1}} = \sqrt {u_{n - 1}^2 + 4{u_{n - 1}}} \\ \Leftrightarrow 4u_n^2 - 4{u_n}{u_{n - 1}} + u_{n - 1}^2 = u_{n - 1}^2 + 4{u_{n - 1}}\\ \Leftrightarrow u_n^2 = \left( {{u_n} + 1} \right){u_{n - 1}}\\ \Rightarrow \frac{1}{{u_n^2}} = \frac{1}{{\left( {{u_n} + 1} \right){u_{n - 1}}}} = \frac{1}{{{u_{n - 1}}}} - \frac{1}{{{u_n}}}\end{array}\)

Vì:

\(\begin{array}{*{20}{l}}{u_n^2 = \left( {{u_n} + 1} \right){u_{n - 1}} = {u_{n - 1}}{u_n} + {u_{n - 1}}}\\{ \Leftrightarrow {u_n} = u_n^2 + {u_n} - {u_{n - 1}}{u_n} - {u_{n - 1}}}\\{ \Leftrightarrow {u_n} = {u_n}\left( {{u_n} + 1} \right) - {u_{n - 1}}\left( {{u_n} + 1} \right)}\\{ \Leftrightarrow {u_n} = \left( {{u_n} + 1} \right)\left( {{u_n} - {u_{n - 1}}} \right)}\\{ \Leftrightarrow \frac{1}{{{u_n} + 1}} = \frac{{{u_n} - {u_{n - 1}}}}{{{u_n}}}}\\{ \Leftrightarrow \frac{1}{{\left( {{u_n} + 1} \right){u_{n - 1}}}} = \frac{{{u_n} - {u_{n - 1}}}}{{{u_n}{u_{n - 1}}}} = \frac{1}{{{u_{n - 1}}}} - \frac{1}{{{u_n}}}}\end{array}\)

Do đó

\(\begin{array}{l}{v_n} = \sum\limits_{i = 1}^n {\frac{1}{{u_i^2}} = \frac{1}{{u_1^2}} + \left( {\frac{1}{{{u_1}}} - \frac{1}{{{u_2}}}} \right) + \left( {\frac{1}{{{u_2}}} - \frac{1}{{{u_3}}}} \right) + ... + \left( {\frac{1}{{{u_{n - 1}}}} - \frac{1}{{{u_n}}}} \right)} \\\,\,\,\,\,\,\, = \frac{1}{{u_1^2}} + \frac{1}{{{u_1}}} - \frac{1}{{{u_n}}} = 6 - \frac{1}{{{u_n}}}\\ \Rightarrow \lim {v_n} = \lim \left( {6 - \frac{1}{{{u_n}}}} \right) = 6 - 0 = 6.\end{array}\)

Chọn D.