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aops MHF1U:
Let $y_{n}=n, \forall n \in \mathbb{N}$.
Since $a_0 > 0$, $a_n$ is a positive series and $a_{n+1} > a_{n}, \forall n \in \mathbb{N}$. This leads to $ \sqrt [k] {{a_{n+1}}^{k+1}} > \sqrt [k] {{a_{n}}^{k+1}}, \forall n \in \mathbb{N}$ or we obtain :
$$ \vert \sqrt [k] {{a_{n+1}}^{k+1}} - \sqrt [k] {{a_{n}}^{k+1}} \vert = \sqrt [k] {{a_{n+1}}^{k+1}} - \sqrt [k] {{a_{n}}^{k+1}} < \sqrt [k] { {a_{n+1}}^{k+1}-{a_{n}}^{k+1}}$$
On the other hand, we have: ${(a_{n+1}-a_{n})}^{k+1}= {a_{n+1}}^{k+1}-{a_{n}}^{k+1} + \sum_{i=1}^{k} C^i_{k+1}(a_{n}a_{n+1})^{i}({a_{n+1}}^{k+1-2i} - {a_{n}}^{k+1-2i}) > {a_{n+1}}^{k+1}-{a_{n}}^{k+1} $.
Moreover, it is obvious that $\displaystyle \lim_{n \to \infty} {a_{n}} = + \infty $ and $a_{n}$ is increasing so $\dfrac {1} {\sqrt [k] {a_n}} \to 0$ when $n \to \infty$. So we have :
$$\sqrt [k] { {a_{n+1}}^{k+1}-{a_{n}}^{k+1}} < \sqrt [k] {{(a_{n+1}-a_{n})}^{k+1}} = \sqrt [k] {\dfrac {1} {\sqrt [k] {a_n}}}^{k+1}= {a_n}^{- \dfrac{k+1}{k^2}} $$Or : $ \vert \dfrac {\sqrt [k] {{a_{n+1}}^{k+1}} - \sqrt [k] {{a_{n}}^{k+1}}}{y_{n+1} - y_n} \vert= \vert \sqrt [k] {{a_{n+1}}^{k+1}} - \sqrt [k] {{a_{n}}^{k+1}} \vert < {a_n}^{- \dfrac{k+1}{k^2}} $
Then $\forall \epsilon > 0, \exists N, \forall n > N \Longrightarrow \vert \dfrac {\sqrt [k] {{a_{n+1}}^{k+1}} - \sqrt [k] {{a_{n}}^{k+1}}}{y_{n+1} - y_n} \vert < {a_n}^{- \dfrac{k+1}{k^2}} < \epsilon$ or $ \lim_{n \to \infty} {\dfrac {\sqrt [k] {{a_{n+1}}^{k+1}} - \sqrt [k] {{a_{n}}^{k+1}}}{y_{n+1} - y_n} } = 0 $
Since all the conditions are satisfied in Stolz's theorem, then we finally have :
$$ \displaystyle \lim_{n \to \infty} {\dfrac {\sqrt [k] {{a_{n+1}}^{k+1} } } {n+1}}
=\displaystyle \lim_{n \to \infty} {\dfrac {\sqrt [k] {{a_{n+1}}^{k+1}} - \sqrt [k] {{a_{n}}^{k+1}}}{y_{n+1} - y_n} } = 0 $$So $ \lim_{n \to \infty} {\dfrac {{a_n}^{k+1} } {n^{k}}} = 0 $ |
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Czhang271828
发表于 2022-7-22 15:56
