Cho dãy số (un) xác định bởi

\[{u_2} = \sqrt {1.2.3.4 + 1} = 5,{u_n} >0,\forall n = 1;2;...\]

Ta có:

\[{u_{n + 1}} = \sqrt {{u_n}({u_n} + 1)({u_n} + 2)({u_n} + 3) + 1} \]

\[ = \sqrt {(u_n^2 + 3{u_n})(u_n^2 + 3{u_n} + 2) + 1} \]

\[ = \sqrt {{{(u_n^2 + 3{u_n})}^2} + 2(u_n^2 + 3{u_n}) + 1} \]

\[ = \sqrt {{{(u_n^2 + 3{u_n} + 1)}^2}} = u_n^2 + 3{u_n} + 1\]

\(\begin{array}{l} \Rightarrow {u_{n + 1}} + 1 = u_n^2 + 3{u_n} + 2 = ({u_n} + 1)({u_n} + 2)\\ \Rightarrow \frac{1}{{{u_{n + 1}} + 1}} = \frac{1}{{({u_n} + 1)({u_n} + 2)}} = \frac{1}{{{u_n} + 1}} - \frac{1}{{{u_n} + 2}}\\ \Rightarrow \frac{1}{{{u_n} + 2}} = \frac{1}{{{u_n} + 1}} - \frac{1}{{{u_{n + 1}} + 1}}\end{array}\)

Do đó: 

\(\begin{array}{l}\\{v_n} = \mathop \sum \limits_{i = 1}^n \frac{1}{{{u_i} + 2}} = \mathop \sum \limits_{i = 1}^n \left( {\frac{1}{{{u_i} + 1}} - \frac{1}{{{u_{i + 1}} + 1}}} \right)\end{array}\)

\[ = \frac{1}{{{u_1} + 1}} - \frac{1}{{{u_{n + 1}} + 1}} = \frac{1}{2} - \frac{1}{{{u_{n + 1}} + 1}}\]

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

\[\lim {u_{n + 1}} = \lim {u_n} = a >0 \Rightarrow a = {a^2} + 3a + 1 \Rightarrow {a^2} + 2a + 1 = 0 \Leftrightarrow a = - 1\,\,\left( {ktm} \right) \Rightarrow \lim {u_n} = + \infty \]\[ \Rightarrow \lim {v_n} = \frac{1}{2} - \frac{1}{{{u_{n + 1}} + 1}} = \frac{1}{2} - 0 = \frac{1}{2}.\]