LimitOfGeometricSequence 2008-11-19 16:51:27 2008-11-19 17:28:37 Proof geometric sequence 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \theoremstyle{definition} \newtheorem*{thmplain}{Theorem} As mentionned in the geometric sequence entry, \begin{align} \lim_{n\to\infty}ar^n = 0 \end{align} for\, $|r| < 1$.\, We will prove this for real or complex values of $r$.\\ We first remark, that for the values\, $s > 1$\, we have\; $\displaystyle\lim_{n\to\infty}s^n = \infty$ (cf. limit of real number sequence).\, In fact, if $M$ is an arbitrary positive number, the binomial theorem (or Bernoulli's inequality) implies that $$s^n = (1+s-1)^n > 1^n+\binom{n}{1}(s-1) = 1+n(s-1) > n(s-1) > M$$ as soon as\, $\displaystyle n > \frac{M}{s-1}$.\\ Let now\, $|r| < 1$\, and $\varepsilon$ be an arbitrarily small positive number.\, Then\, $\displaystyle|r| = \frac{1}{s}$\, with\,$s > 1$.\, By the above remark, $$|r^n| = |r|^n = \frac{1}{s^n} < \frac{1}{n(s-1)} < \varepsilon$$ when\, $\displaystyle n > \frac{1}{(s-1)\varepsilon}$.\, Hence, $$\lim_{n\to\infty}r^n =0,$$ which easily implies (1) for any real number $a$.