Lambert series LambertSeries 2009-01-26 08:53:19 2009-01-31 18:01:29 Example double series % 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} \PMlinkescapeword{terms} The series \begin{align} \sum_{n=1}^\infty\frac{a_nz^n}{1\!-\!z^n} \;=\; \frac{a_1z}{1\!-\!z}+\frac{a_2z^2}{1\!-\!z^2}+\ldots \end{align} is called {\em Lambert series}.\, We here consider more closely only the special case \begin{align} \sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; \frac{x}{1\!-\!x}+\frac{x^2}{1\!-\!x^2}+\ldots \end{align} for the real \PMlinkescapetext{variable} $x$.\\ \textbf{I.\; Convergence}\\ $1^\circ.$\; $x = \pm1$:\; The series is not defined.\\ $2^\circ.$\; $|x| > 1$:\; We have \[ \frac{x^n}{1\!-\!x^n} \;=\; \frac{1}{\frac{1}{x^n}\!-\!1} \;\to\; -1 \neq 0 \quad \mbox{as}\;\; n \to \infty, \] whence the series (2) diverges.\\ $3^\circ.$\; $0 \leqq x < 1$:\; The series with nonnegative terms converges, since \[ \sqrt[n]{\frac{x^n}{1\!-\!x^n}} \;=\; \frac{x}{\sqrt[n]{1\!-\!x^n}} \;\to\; x < 1 \quad \mbox{as}\;\; n \to \infty. \] $4^\circ.$\; $-1 < x < 0$:\; We get an alternating series with \[ \left|\frac{x^n}{1\!-\!x^n}\right| \;=\; \frac{|x|^n}{|1\!-\!x^n|} \;\leqq\; \frac{|x|^n}{1\!-\!|x|^n} \;\leqq\; \frac{|x|^n}{1\!-\!|x|} \;\to\; 0 \quad \mbox{as}\;\; n \to \infty, \] and by Leibniz theorem, the series converges.\\ Thus we have the result that the Lambert series (2) converges, \PMlinkescapetext{even} absolutely, when\, $|x| < 1$.\\ \PMlinkescapetext{\textbf{II.\; Power series expansion}}\\ Let\, $|x| < 1$.\, \PMlinkescapetext{Expand} the terms to geometric series:\\ \begin{tabular}{lcrrrrrrrrrrrrrrr} $\displaystyle\frac{x}{1\!-\!x}$&$=$&$x$&$+$&$x^2$&$+$&$x^3$&$+$&$x^4$&$+$&$x^5$&$+$&$x^6$&$+$&$\ldots$\\ $\displaystyle\frac{x^2}{1\!-\!x^2}$&$=$&$$&$$&$x^2$&$$&$$&$+$&$x^4$&$$&$$&$+$&$x^6$&$+$&$\ldots$\\ $\displaystyle\frac{x^3}{1\!-\!x^3}$&$=$&$$&$$&$$&$$&$x^3$&$$&$$&$$&$$&$+$&$x^6$&$+$&$\ldots$\\ $\displaystyle\frac{x^4}{1\!-\!x^4}$&$=$&$$&$$&$$&$$&$$&$$&$x^4$&$$&$$&$$&$$&$+$&$\ldots$\\ $\displaystyle\frac{x^5}{1\!-\!x^5}$&$=$&$$&$$&$$&$$&$$&$$&$$&$$&$x^5$&$$&$$&$+$&$\ldots$\\ $\displaystyle\frac{x^6}{1\!-\!x^6}$&$=$&$$&$$&$$&$$&$$&$$&$$&$$&$$&$$&$x^6$&$+$&$\ldots$\\ $\displaystyle\;\;\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$\\ \end{tabular} Those geometric series converge absolutely, \[ |x^k|+|x^{2k}|+|x^{3k}|+\ldots \;=\; \frac{|x|^k}{1\!-\!|x|^k} \] and the series $\displaystyle\sum_{k=1}^\infty\frac{|x|^k}{1\!-\!|x|^k}$ converges.\, Thus we can sum the geometric series by the columns: \[ \sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; x+2x^2+2x^3+3x^4+2x^5+4x^6+\ldots \] Apparently, the coefficient of any $x^k$ in this power series expresses, by how many positive integers the number $k$ is divisible, i.e. the coefficient is given by the divisor function $\tau$.\, So we may write the power series form of the Lambert series as \[ \sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; \tau(1)x+\tau(2)x^2+\tau(3)x^3+\ldots \]