Cho dãy số $({u_n})$xác định bởi
\(\left\{ \begin{array}{l}{u_1} = \dfrac{1}{2}\\{u_{n + 1}} = \dfrac{{\sqrt {u_n^2 + 4{u_n}} + {u_n}}}{2},\left( {n \ge 1} \right)\end{array} \right.\)
Đặt ${v_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{u_{_i}^2}}}, $ khẳng định nào sau đây đúng?

Xét \({u_{n + 1}} - {u_n} = \dfrac{{\sqrt {u_n^2 + 4{u_n}} + {u_n}}}{2} - {u_n} \)

\(= \dfrac{{\sqrt {u_n^2 + 4{u_n}} - {u_n}}}{2} > \dfrac{{\sqrt {u_n^2} - {u_n}}}{2} = 0 \)

\(\Rightarrow \left( {{u_n}} \right)\) là dãy tăng.
Giả sử $\lim {u_n} = a$ thì $a > 0$ và $a = \dfrac{{\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 $
\(\begin{array}{l}{u_n} = \dfrac{{\sqrt {u_{n - 1}^2 + 4{u_{n - 1}}} + {u_{n - 1}}}}{2} \\ \Leftrightarrow 2{u_n} - {u_{n - 1}} = \sqrt {u_{n - 1}^2 + 4{u_{n - 1}}} \\ \Leftrightarrow 4u_n^2 - 4{u_n}{u_{n - 1}} + u_{n - 1}^2 = u_{n - 1}^2 + 4{u_{n - 1}} \\ \Leftrightarrow u_n^2 = \left( {{u_n} + 1} \right){u_{n - 1}}\end{array}\)

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