Lambert series

The series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{a_n{z}^n}{1-z^n}}={\displaystyle \frac{a_{1}z}{1-z}}+{\displaystyle \frac{a_2{z}^2}{1-z^2}}+\mathrm{\dots }$ (1)

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

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{x^n}{1-x^n}}={\displaystyle \frac{x}{1-x}}+{\displaystyle \frac{x^2}{1-x^2}}+\mathrm{\dots }$ (2)

for the real $x$.

I.  Convergence

$1^{\circ }.$  $x=\pm 1$:  The series is not defined.

$2^{\circ }.$  $|x|>1$:  We have

$$\frac{x^n}{1-x^n}=\frac{1}{\frac{1}{x^n}-1}\to -1\ne 0\mathit{\hspace{1em}}\text{as}n\to \mathrm{\infty },$$

whence the series (2) diverges.

$3^{\circ }.$  :  The series with nonnegative terms converges, since

$4^{\circ }.$  :  We get an alternating series with

$$\left|\frac{x^n}{1-x^n}\right|=\frac{{|x|}^n}{|1-x^n|}\leqq \frac{{|x|}^n}{1-{|x|}^n}\leqq \frac{{|x|}^n}{1-|x|}\to \mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\text{as}n\to \mathrm{\infty },$$

and by Leibniz theorem, the series converges.

Thus we have the result that the Lambert series (2) converges, absolutely, when  .

Let  .  the terms to geometric series:
${\displaystyle \frac{x}{1-x}}$ $=$ $x$ $+$ $x^2$ $+$ $x^3$ $+$ $x^4$ $+$ $x^5$ $+$ $x^6$ $+$ $\mathrm{\dots }$ ${\displaystyle \frac{x^2}{1-x^2}}$ $=$ $x^2$ $+$ $x^4$ $+$ $x^6$ $+$ $\mathrm{\dots }$ ${\displaystyle \frac{x^3}{1-x^3}}$ $=$ $x^3$ $+$ $x^6$ $+$ $\mathrm{\dots }$ ${\displaystyle \frac{x^4}{1-x^4}}$ $=$ $x^4$ $+$ $\mathrm{\dots }$ ${\displaystyle \frac{x^5}{1-x^5}}$ $=$ $x^5$ $+$ $\mathrm{\dots }$ ${\displaystyle \frac{x^6}{1-x^6}}$ $=$ $x^6$ $+$ $\mathrm{\dots }$ $\mathrm{\dots }$ $\mathrm{\dots }$ $\mathrm{\dots }$ $\mathrm{\dots }$ $\mathrm{\dots }$ $\mathrm{\dots }$ $\mathrm{\dots }$ $\mathrm{\dots }$

Those geometric series converge absolutely,

$$|x^k|+|x^{2k}|+|x^{3k}|+\mathrm{\dots }=\frac{{|x|}^k}{1-{|x|}^k}$$

and the series ${\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{{|x|}^k}{1-{|x|}^k}}$ converges.  Thus we can sum the geometric series by the columns:

$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{1-x^n}=x+2x^2+2x^3+3x^4+2x^5+4x^6+\mathrm{\dots }$$

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

$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{1-x^n}=\tau (1)x+\tau (2)x^2+\tau (3)x^3+\mathrm{\dots }$$

Title	Lambert series
Canonical name	LambertSeries
Date of creation	2013-03-22 18:46:42
Last modified on	2013-03-22 18:46:42
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	13
Author	pahio (2872)
Entry type	Example
Classification	msc 30B10
Classification	msc 40A05
Related topic	NecessaryConditionOfConvergence
Related topic	CauchysRootTest
Related topic	TauFunction