\(\text { Cho dãy số }\left(\mathrm{u}_{\mathrm{n}}\right) \text { được xác định bởi: }\left\{\begin{array}{l} \mathrm{u}_{0}=2011 \\ \mathrm{u}_{\mathrm{n}+1}=\mathrm{u}_{\mathrm{n}}+\frac{1}{\mathrm{u}_{\mathrm{n}}^{2}} \end{array} \text { . Tìm } \lim \frac{\mathrm{u}_{\mathrm{n}}^{3}}{\mathrm{n}}\right. \text { . }\)
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Lời giải:
Báo sai\(\begin{aligned} &\text { T. thấy } \mathrm{u}_{\mathrm{n}}>0, \forall \mathrm{n}\\ &\text { Ta có: } u_{n+1}^{3}=u_{n}^{3}+3+\frac{3}{u_{n}^{3}}+\frac{1}{u_{n}^{6}}(1)\\ &\text { Suy ra: } \mathrm{u}_{\mathrm{n}}^{3}>\mathrm{u}_{\mathrm{n}-1}^{3}+3 \Rightarrow \mathrm{u}_{\mathrm{n}}^{3}>\mathrm{u}_{0}^{3}+3 \mathrm{n}(2)\\ &\text { Từ (1) và (2), suy ra: } \mathrm{u}_{\mathrm{n}+1}^{3}<\mathrm{u}_{\mathrm{n}}^{3}+3+\frac{1}{\mathrm{u}_{0}^{3}+3 \mathrm{n}}+\frac{1}{\left(\mathrm{u}_{0}^{3}+3 \mathrm{n}\right)^{2}}<\mathrm{u}_{\mathrm{n}}^{3}+3+\frac{1}{3 \mathrm{n}}+\frac{1}{9 \mathrm{n}^{2}} \end{aligned}\)
\(\begin{array}{l} \text { Do đó: } \mathrm{u}_{\mathrm{n}}^{3}<\mathrm{u}_{0}^{3}+3 \mathrm{n}+\frac{1}{3} \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}}+\frac{1}{9} \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}^{2}} \text { (3) } \\ \text { Lại có: } \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}^{2}}<1+\frac{1}{1.2}+\frac{1}{2.3}+\ldots+\frac{1}{(\mathrm{n}-1) \mathrm{n}}=2-\frac{1}{\mathrm{n}}<2 \\ \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}} \leq \sqrt{\mathrm{n}} \sqrt{\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}^{2}}}<\sqrt{2 \mathrm{n}} \\ \text { Nên: } \mathrm{u}_{0}^{3}+3 \mathrm{n}<\mathrm{u}_{\mathrm{n}}^{3}<\mathrm{u}_{0}^{3}+3 \mathrm{n}+\frac{2}{9}+\frac{\sqrt{2 \mathrm{n}}}{3} \\ \text { Hay } 3+\frac{\mathrm{u}_{0}^{3}}{\mathrm{n}}<\frac{\mathrm{u}_{\mathrm{n}}^{3}}{\mathrm{n}}<3+\frac{\mathrm{u}_{0}^{3}}{\mathrm{n}}+\frac{2}{9 \mathrm{n}}+\frac{\sqrt{2}}{3 \sqrt{\mathrm{n}}} \end{array}\)
\(\text { Vậy } \lim \frac{\mathrm{u}_{\mathrm{n}}^{3}}{\mathrm{n}}=3\)
