Given ${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } }$ $\therefore$ ${a_n} = \sqrt {7 + {a_n}}$ $\Rightarrow$ $a_n^2 = 7 + {a_n}$ $\Rightarrow$ $a_n^2 - {a_n} - 7 = 0$ $\Rightarrow {a_n} = {{1 \pm \sqrt {1 - 4 \times 1 \times - 7} } \over 2}$ $\Rightarrow {a_n} = {{1 \pm \sqrt {29} } \over 2}$ As ${a_n}$ > 0, $\therefore$ ${a_n} = {{1 + \sqrt {29} } \over 2}$ = 3.19 So ${a_n} > 3\,\,\forall \,\,n \ge 1$