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

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