Cho dãy số thực ()
$ \left\{ \begin{array}{l} a_1 = 1 \\ a_{n + 1} = a_n + \dfrac{1}{{a_n }}(n \ge 1) \\ \end{array} \right. $
Chứng minh:
$\lim \dfrac{{a_n }}{{\sqrt n }} = \sqrt 2 $
mình chém luôn nha
Ta có
$a_{k + 1}^2 = a_k^2 + \dfrac{1}{{a_k^2}} + 2 \Rightarrow \sum\limits_{1 = 1}^n {a_i^2} = \sum\limits_{j = 1}^{n - 1} {a_j^2} + \sum\limits_{j = 1}^{n - 1} {\dfrac{1}{{a_j^2}}} + 2(n - 1) \to a_n^2 = 2n - 1 + \sum\limits_{j = 1}^{n - 1} {\dfrac{1}{{a_j^2}}} \to {a_n} > \sqrt {2n - 1} \forall n \ge 2 \to a_k^2 > 2k - 1,\forall k \ge 2 \Rightarrow \dfrac{1}{{a_k^4}} < \dfrac{1}{{{{(2k - 1)}^2}}} < \dfrac{1}{{4k(k - 1)}} = \dfrac{1}{4}(\dfrac{1}{{k - 1}} - \dfrac{1}{k}) \to \sum\limits_{k = 2}^{n - 1} {\dfrac{1}{{a_k^4}}} < \dfrac{1}{4}(1 - \dfrac{1}{{n - 1}}) < \dfrac{1}{4} \to \sum\limits_{k = 1}^{n - 1} {\dfrac{1}{{a_k^4}}} < \dfrac{5}{4} \to \sum\limits_{j = 1}^{n - 1} {\dfrac{1}{{a_j^2}}} \le \sqrt {(n - 1)\sum\limits_{k = 2}^{n - 1} {\dfrac{1}{{a_k^4}}} } < \sqrt {(n - 1)\dfrac{5}{4}} \to a_n^2 < 2n - 1 + \sqrt {(n - 1)\dfrac{5}{4}} $
$\forall n \ge 2,\sqrt {2n - 1} < {a_n} < \sqrt {2n - 1 + \sqrt {(n - 1)\dfrac{5}{4}} } \to \sqrt {2 - \dfrac{1}{n}} < \dfrac{{{a_n}}}{{\sqrt n }} < \sqrt {2 - \dfrac{1}{n} + \sqrt {(\dfrac{1}{n} - \dfrac{1}{{{n^2}}})\dfrac{5}{4}} } \to \mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_n}}}{{\sqrt n }} = \sqrt 2 $