Αναδρομική ακολουθία (3)

#1

Έστω $\displaystyle{x_{n+1}=x_{n}+x_{n}^{-1/k}}$ με $k\in\mathbb N$ και $x_{0}>0$ . Βρείτε τα

$\displaystyle{\lim_{n\to+\infty}\frac{x_{n}^{k+1}}{n^k}}$ ,

$\displaystyle{\lim_{n\to+\infty}\left[\frac{x_{n}^{k+1}}{n^k}-\left(1+\frac{1}{k}\right)^k\right]\frac{n}{\ln n}}$ .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

#2

Κοτρώνης Αναστάσιος έγραψε:Έστω $\displaystyle{x_{n+1}=x_{n}+x_{n}^{-1/k}}$ με $k\in\mathbb N$ και $x_{0}>0$ . Βρείτε τα

$\displaystyle{\lim_{n\to+\infty}\frac{x_{n}^{k+1}}{n^k}}$ ,

$\displaystyle{\lim_{n\to+\infty}\left[\frac{x_{n}^{k+1}}{n^k}-\left(1+\frac{1}{k}\right)^k\right]\frac{n}{\ln n}}$ .

Για το πρώτο $\displaystyle{{x_{n + 1}} = {x_n} + \frac{1}{{{{\left( {{x_n}} \right)}^{1/k}}}} \Rightarrow {\left( {{x_{n + 1}}} \right)^{\left( {k + 1} \right)/k}} - {\left( {{x_n}} \right)^{\left( {k + 1} \right)/k}} = {\left( {{x_n} + \frac{1}{{{{\left( {{x_n}} \right)}^{1/k}}}}} \right)^{\left( {k + 1} \right)/k}} - {\left( {{x_n}} \right)^{\left( {k + 1} \right)/k}} = }$

$\displaystyle{ = {\left( {{x_n}} \right)^{\left( {k + 1} \right)/k}}\left( {{{\left( {1 + \frac{1}{{x_n^{\left( {k + 1} \right)/k}}}} \right)}^{\left( {k + 1} \right)/k}} - 1} \right) = {\left( {{x_n}} \right)^{\left( {k + 1} \right)/k}}\left( {\left( {1 + \frac{{k + 1}}{k} \cdot \frac{1}{{x_n^{\left( {k + 1} \right)/k}}} + O\left( {\frac{1}{{x_n^{2\left( {k + 1} \right)/k}}}} \right)} \right) - 1} \right) = }$

$\displaystyle{ = \frac{{k + 1}}{k} + O\left( {\frac{1}{{x_n^{\left( {k + 1} \right)/k}}}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {{x_{n + 1}}} \right)}^{\left( {k + 1} \right)/k}} - {{\left( {{x_n}} \right)}^{\left( {k + 1} \right)/k}}}}{{\left( {n + 1} \right) - n}}} \right) = \frac{{k + 1}}{k} \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {{x_n}} \right)}^{\left( {k + 1} \right)/k}}}}{n} = \frac{{k + 1}}{k}}$ (Stolz –Cesaro).

Επομένως $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \frac{{x_n^{k + 1}}}{{{n^k}}} = {\left( {\frac{{k + 1}}{k}} \right)^k}}$

Επίσης ισχύει $\displaystyle{{x_n} \sim {\left( {\frac{{k + 1}}{k} \cdot n} \right)^{k/(k + 1)}}}$

Για το δεύτερο έχουμε $\displaystyle{{\left( {{x_{n + 1}}} \right)^{\left( {k + 1} \right)/k}} - {\left( {{x_n}} \right)^{\left( {k + 1} \right)/k}} = \frac{{k + 1}}{k} + \frac{1}{{x_n^{\left( {k + 1} \right)/k}}} + O\left( {\frac{1}{{x_n^{2\left( {k + 1} \right)/k}}}} \right) \Rightarrow }$ (επειδή $\displaystyle{{x_n} \sim {\left( {\frac{{k + 1}}{k} \cdot n} \right)^{k/(k + 1)}}}$ )

$\displaystyle{ \Rightarrow {\left( {{x_{n + 1}}} \right)^{\left( {k + 1} \right)/k}} - {\left( {{x_n}} \right)^{\left( {k + 1} \right)/k}} - \frac{{k + 1}}{k} \sim \frac{k}{{\left( {k + 1} \right)n}} + O\left( {\frac{1}{{x_n^{2\left( {k + 1} \right)/k}}}} \right) \Rightarrow }$

$\displaystyle{ \Rightarrow \sum\limits_{j = 1}^n {\left( {{{\left( {{x_{j + 1}}} \right)}^{\left( {k + 1} \right)/k}} - {{\left( {{x_j}} \right)}^{\left( {k + 1} \right)/k}} - \frac{{k + 1}}{k}} \right)} \sim \frac{k}{{k + 1}}\sum\limits_{j = 1}^n {\frac{1}{j}} \sim \frac{{k \cdot \ln n}}{{k + 1}} \Rightarrow }$

$\displaystyle{ \Rightarrow {\left( {{x_{n + 1}}} \right)^{\left( {k + 1} \right)/k}} - {\left( {{x_1}} \right)^{\left( {k + 1} \right)/k}} - \frac{{\left( {k + 1} \right) \cdot n}}{k} \sim \frac{{k \cdot \ln n}}{{k + 1}} \Rightarrow {\left( {{x_{n + 1}}} \right)^{\left( {k + 1} \right)/k}} \sim \frac{{k + 1}}{k} \cdot n + \frac{k}{{k + 1}} \cdot \ln n \Rightarrow }$

$\displaystyle{ \Rightarrow {\left( {{x_{n + 1}}} \right)^{\left( {k + 1} \right)}} \sim {\left( {\frac{{k + 1}}{k} \cdot n + \frac{k}{{k + 1}} \cdot \ln n} \right)^k} \Rightarrow \frac{{{{\left( {{x_{n + 1}}} \right)}^{\left( {k + 1} \right)}}}}{{{n^k}}} \sim {\left( {\frac{{k + 1}}{k} + \frac{k}{{k + 1}} \cdot \frac{{\ln n}}{n}} \right)^k} \Rightarrow }$

$\displaystyle{ \Rightarrow \frac{{{{\left( {{x_{n + 1}}} \right)}^{\left( {k + 1} \right)}}}}{{{n^k}}} = {\left( {\frac{{k + 1}}{k}} \right)^k} + k{\left( {\frac{{k + 1}}{k}} \right)^{k - 1}}\frac{k}{{k + 1}} \cdot \frac{{\ln n}}{n} + O\left( {\frac{{{{\ln }^2}n}}{{{n^2}}}} \right) \Rightarrow }$

$\displaystyle{ \Rightarrow \frac{{{{\left( {{x_{n + 1}}} \right)}^{\left( {k + 1} \right)}}}}{{{n^k}}} - {\left( {\frac{{k + 1}}{k}} \right)^k} = k{\left( {\frac{{k + 1}}{k}} \right)^{k - 2}} \cdot \frac{{\ln n}}{n} + O\left( {\frac{{{{\ln }^2}n}}{{{n^2}}}} \right) \Rightarrow }$

$\displaystyle{ \Rightarrow \left( {\frac{{{{\left( {{x_{n + 1}}} \right)}^{\left( {k + 1} \right)}}}}{{{n^k}}} - {{\left( {\frac{{k + 1}}{k}} \right)}^k}} \right)\frac{n}{{\ln n}} = k{\left( {\frac{{k + 1}}{k}} \right)^{k - 2}} + O\left( {\frac{{\ln n}}{n}} \right)}$

Τελικά $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {{x_{n + 1}}} \right)}^{\left( {k + 1} \right)}}}}{{{n^k}}} - {{\left( {\frac{{k + 1}}{k}} \right)}^k}} \right)\frac{n}{{\ln n}} = k{\left( {\frac{{k + 1}}{k}} \right)^{k - 2}}}$

Αναδρομική ακολουθία (3)

Αναδρομική ακολουθία (3)

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