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40-01-1fracalphannIsMonotoneForLargeN.tex
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40-01-1fracalphannIsMonotoneForLargeN.tex
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\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{1fracalphannIsMonotoneForLargeN}
\pmcreated{2013-03-22 17:53:55}
\pmmodified{2013-03-22 17:53:55}
\pmowner{uriw}{288}
\pmmodifier{uriw}{288}
\pmtitle{$(1+\frac{\alpha}{n})^n$ is monotone for large $n$}
\pmrecord{4}{40389}
\pmprivacy{1}
\pmauthor{uriw}{288}
\pmtype{Theorem}
\pmcomment{trigger rebuild}
\pmclassification{msc}{40-01}
\pmclassification{msc}{00-01}
\endmetadata
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}
\newtheorem*{lemma}{Lemma}
\begin{document}
\begin{lemma}
Let $\alpha$ be a real number. The sequence
$(1+\frac{\alpha}{n})^n$ is monotone increasing for all $n>
|\alpha|$.
\end{lemma}
\begin{proof}
Let $n>|\alpha|$. We want to prove the following inequality:
\[
\left(1+\frac{\alpha}{n}\right)^n \leq
\left(1+\frac{\alpha}{n+1}\right)^{n+1}
\]
Since both sides are positive, this follows by taking the
$(n+1)$-th root and using the arithmetic-geometric-harmonic means
inequality:
\[
\sqrt[n+1]{\left(1+\frac{\alpha}{n}\right)^n} =
\underbrace{\sqrt[n+1]{1\cdot \left(1+\frac{\alpha}{n}\right)
\cdots \left(1+\frac{\alpha}{n}\right)} }_{\textrm{$n+1$ elements}
} \leq \frac{1+n\left(1+\frac{\alpha}{n}\right)}{n+1} =
1+\frac{\alpha}{n+1}
\]
\end{proof}
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\end{document}