% Priklad do sbirky, kapitola 12, tezky priklad c. 1
% Napsano v LaTeXu

\documentclass[a4paper,12pt]{article}

\usepackage[cp852]{inputenc}
\usepackage[czech]{babel}
\usepackage{amssymb}

\newcommand{\en}{\mathbb N}

\newcommand{\implies}{\Longrightarrow}

\newcommand{\eps}{\varepsilon}

\newcommand{\lm}{\lim_{n \to \infty}}
\newcommand{\ls}{\limsup_{n \to \infty}}
\newcommand{\li}{\liminf_{n \to \infty}}

\newcommand{\seq}[2]{(#1_#2)_{#2=1}^{+\infty}}


\begin{document}

Nechœ $\seq{a}{n}$ je posloupnost kladnìch re lnìch Ÿ¡sel. OznaŸme
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\[ A = \ls \frac{a_{n+1}}{a_n} \]
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Zvolme $\eps > 0$ libovoln‚. Potom existuje $n_0 \in \en$ takov‚, §e
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\[ n \ge n_0 \implies \frac{a_{n+1}}{a_n} < A + \frac{\eps}{2} \]
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Indukc¡ snadno dok §eme, §e pro ka§d‚ $k \in \en$ je
$a_{n_0+k} < a_{n_0} \, (A + \eps/2)^k$. Zvolme Ÿ¡slo $K$ takto:
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\[ K = \max \left\{ \frac{a_1}{(A+\eps/2)}, \frac{a_2}{(A+\eps/2)^2}, \dots,
   \frac{a_{n_0}}{(A+\eps/2)^{n_0}} \right\} \]
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Potom pro ka§d‚ $n$ pýirozen‚ plat¡ $a_n \le K \, (A + \eps/2)^n$, a tedy
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\[ \sqrt[n]{a_n} \le \sqrt[n]{K} \, \left( A + \frac{\eps}{2} \right) \]
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Proto§e $\lm \sqrt[n]{K} \, (A + \eps/2) = (A + \eps/2)$, existuje $m \in \en$ takov‚, §e
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\[ n > m \implies \sqrt[n]{a_n} < A + \eps \]
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To znamen , §e
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\[ \ls \sqrt[n]{a_n} \le A + \eps \]
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Proto§e jsme $\eps$ zvolili jako \emph{libovoln‚} kladn‚ Ÿ¡slo, plat¡:
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\[ \ls \sqrt[n]{a_n} \le A = \ls \frac{a_{n+1}}{a_n} \]
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Nerovnost $\li \sqrt[n]{a_n} \le \ls \sqrt[n]{a_n}$ je zýejm .
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Pokud
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\[ \li \frac{a_{n+1}}{a_n} = 0 \]
je nerovnost 
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\[ \li \frac{a_{n+1}}{a_n} \le \li \sqrt[n]{a_n} \]
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zýejm . Jinak ji dok §eme analogicky jako
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\[ \ls \sqrt[n]{a_n} \le \ls \frac{a_{n+1}}{a_n} \]
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pouze s t¡m rozd¡lem, §e $\eps$ nem…§eme volit libovolnØ, ale tak,
aby
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\[ \li \frac{a_{n+1}}{a_n} - \eps > 0 \]

\end{document}