# Lambert series

The series

 $\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}z^{n}}{1\!-\!z^{n}}\;=\;\frac{a_{1}% z}{1\!-\!z}+\frac{a_{2}z^{2}}{1\!-\!z^{2}}+\ldots$ (1)

is called .  We here consider more closely only the special case

 $\displaystyle\sum_{n=1}^{\infty}\frac{x^{n}}{1\!-\!x^{n}}\;=\;\frac{x}{1\!-\!x% }+\frac{x^{2}}{1\!-\!x^{2}}+\ldots$ (2)

for the real $x$.

$1^{\circ}.$$x=\pm 1$:  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:  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, absolutely, when  $|x|<1$.

Let  $|x|<1$.  the terms to geometric series  :
$\displaystyle\frac{x}{1\!-\!x}$ $=$ $x$ $+$ $x^{2}$ $+$ $x^{3}$ $+$ $x^{4}$ $+$ $x^{5}$ $+$ $x^{6}$ $+$ $\ldots$ $\displaystyle\frac{x^{2}}{1\!-\!x^{2}}$ $=$ $x^{2}$ $+$ $x^{4}$ $+$ $x^{6}$ $+$ $\ldots$ $\displaystyle\frac{x^{3}}{1\!-\!x^{3}}$ $=$ $x^{3}$ $+$ $x^{6}$ $+$ $\ldots$ $\displaystyle\frac{x^{4}}{1\!-\!x^{4}}$ $=$ $x^{4}$ $+$ $\ldots$ $\displaystyle\frac{x^{5}}{1\!-\!x^{5}}$ $=$ $x^{5}$ $+$ $\ldots$ $\displaystyle\frac{x^{6}}{1\!-\!x^{6}}$ $=$ $x^{6}$ $+$ $\ldots$ $\displaystyle\;\;\ldots$ $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\ldots$

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

 $\sum_{n=1}^{\infty}\frac{x^{n}}{1\!-\!x^{n}}\;=\;\tau(1)x+\tau(2)x^{2}+\tau(3)% x^{3}+\ldots$
Title Lambert series LambertSeries 2013-03-22 18:46:42 2013-03-22 18:46:42 pahio (2872) pahio (2872) 13 pahio (2872) Example msc 30B10 msc 40A05 NecessaryConditionOfConvergence CauchysRootTest TauFunction