| Version 8 |
Version 7 |
| \PMlinkescapeword{terms} |
\PMlinkescapeword{terms} |
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| The series |
The series |
| \begin{align} |
\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 |
\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} |
\end{align} |
| is called {\em Lambert series}.\, We here consider more closely only the special case |
is called {\em Lambert series}.\, We here consider more closely only the special case |
| \begin{align} |
\begin{align} |
| \sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; \frac{x}{1\!-\!x}+\frac{x^2}{1\!-\!x^2}+\ldots |
\sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; \frac{x}{1\!-\!x}+\frac{x^2}{1\!-\!x^2}+\ldots |
| \end{align} |
\end{align} |
| for the real \PMlinkescapetext{variable} $x$.\\ |
for the real \PMlinkescapetext{variable} $x$.\\ |
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| \textbf{I.\; Convergence}\\ |
\textbf{I.\; Convergence}\\ |
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| $1^\circ.$\; $x = \pm1$:\; The series is not defined.\\ |
$1^\circ.$\; $x = \pm1$:\; The series is not defined.\\ |
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| $2^\circ.$\; $|x| > 1$:\; We have |
$2^\circ.$\; $|x| > 1$:\; We have |
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| \[ |
\[ |
| \frac{x^n}{1\!-\!x^n} \;=\; \frac{1}{\frac{1}{x^n}\!-\!1} \;\to\; -1 \neq 0 |
\frac{x^n}{1\!-\!x^n} \;=\; \frac{1}{\frac{1}{x^n}\!-\!1} \;\to\; -1 \neq 0 |
| \quad \mbox{as}\;\; n \to \infty, |
\quad \mbox{as}\;\; n \to \infty, |
| \] |
\] |
| whence the series (2) diverges.\\ |
whence the series (2) diverges.\\ |
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| $3^\circ.$\; $0 \leqq x < 1$:\; The series with nonnegative terms converges, since |
$3^\circ.$\; $0 \leqq x < 1$:\; The series with nonnegative terms converges, since |
| \[ |
\[ |
| \sqrt[n]{\frac{x^n}{1\!-\!x^n}} \;=\; \frac{x}{\sqrt[n]{1\!-\!x^n}} \;\to\; x < 1 |
\sqrt[n]{\frac{x^n}{1\!-\!x^n}} \;=\; \frac{x}{\sqrt[n]{1\!-\!x^n}} \;\to\; x < 1 |
| \quad \mbox{as}\;\; n \to \infty. |
\quad \mbox{as}\;\; n \to \infty. |
| \] |
\] |
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| $4^\circ.$\; $-1 < x < 0$:\; We get an alternating series with |
$4^\circ.$\; $-1 < x < 0$:\; We get an alternating series with |
| \[ |
\[ |
| \left|\frac{x^n}{1\!-\!x^n}\right| \;=\; \frac{|x|^n}{|1\!-\!x^n|} \;\leqq\; \frac{|x|^n}{1\!-\!|x|^n} |
\left|\frac{x^n}{1\!-\!x^n}\right| \;=\; \frac{|x|^n}{|1\!-\!x^n|} \;\leqq\; \frac{|x|^n}{1\!-\!|x|^n} |
| \;\leqq\; \frac{|x|^n}{1\!-\!|x|} \;\to\; 0 \quad \mbox{as}\;\; n \to \infty, |
\;\leqq\; \frac{|x|^n}{1\!-\!|x|} \;\to\; 0 \quad \mbox{as}\;\; n \to \infty, |
| \] |
\] |
| and by Leibniz theorem, the series converges.\\ |
and by Leibniz theorem, the series converges.\\ |
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| Thus we have the result that the Lambert series (2) converges, \PMlinkescapetext{even} absolutely, when\, $|x| < 1$.\\ |
Thus we have the result that the Lambert series (2) converges, \PMlinkescapetext{even} absolutely, when\, $|x| < 1$.\\ |
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| \PMlinkescapetext{\textbf{II.\; Power series expansion}}\\ |
\PMlinkescapetext{\textbf{II.\; Power series expansion}}\\ |
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| Let\, $|x| < 1$.\, \PMlinkescapetext{Expand} the terms to geometric series:\\ |
Let\, $|x| < 1$.\, \PMlinkescapetext{Expand} the terms to geometric series:\\ |
| \begin{tabular}{lcrrrrrrrrrrrrrrr} |
\begin{tabular}{lcrrrrrrrrrrrrrrr} |
| $\displaystyle\frac{x}{1\!-\!x}$&$=$&$x$&$+$&$x^2$&$+$&$x^3$&$+$&$x^4$&$+$&$x^5$&$+$&$x^6$&$+$&$\ldots$\\ |
$\displaystyle\frac{x}{1\!-\!x}$&$=$&$x$&$+$&$x^2$&$+$&$x^3$&$+$&$x^4$&$+$&$x^5$&$+$&$x^6$&$+$&$\ldots$\\ |
| $\displaystyle\frac{x^2}{1\!-\!x^2}$&$=$&$$&$$&$x^2$&$$&$$&$+$&$x^4$&$$&$$&$+$&$x^6$&$+$&$\ldots$\\ |
$\displaystyle\frac{x^2}{1\!-\!x^2}$&$=$&$$&$$&$x^2$&$$&$$&$+$&$x^4$&$$&$$&$+$&$x^6$&$+$&$\ldots$\\ |
| $\displaystyle\frac{x^3}{1\!-\!x^3}$&$=$&$$&$$&$$&$$&$x^3$&$$&$$&$$&$$&$+$&$x^6$&$+$&$\ldots$\\ |
$\displaystyle\frac{x^3}{1\!-\!x^3}$&$=$&$$&$$&$$&$$&$x^3$&$$&$$&$$&$$&$+$&$x^6$&$+$&$\ldots$\\ |
| $\displaystyle\frac{x^4}{1\!-\!x^4}$&$=$&$$&$$&$$&$$&$$&$$&$x^4$&$$&$$&$$&$$&$+$&$\ldots$\\ |
$\displaystyle\frac{x^4}{1\!-\!x^4}$&$=$&$$&$$&$$&$$&$$&$$&$x^4$&$$&$$&$$&$$&$+$&$\ldots$\\ |
| $\displaystyle\frac{x^5}{1\!-\!x^5}$&$=$&$$&$$&$$&$$&$$&$$&$$&$$&$x^5$&$$&$$&$+$&$\ldots$\\ |
$\displaystyle\frac{x^5}{1\!-\!x^5}$&$=$&$$&$$&$$&$$&$$&$$&$$&$$&$x^5$&$$&$$&$+$&$\ldots$\\ |
| $\displaystyle\frac{x^5}{1\!-\!x^5}$&$=$&$$&$$&$$&$$&$$&$$&$$&$$&$$&$$&$x^6$&$+$&$\ldots$\\ |
$\displaystyle\frac{x^5}{1\!-\!x^5}$&$=$&$$&$$&$$&$$&$$&$$&$$&$$&$$&$$&$x^6$&$+$&$\ldots$\\ |
| $\displaystyle\;\;\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$\\ |
$\displaystyle\;\;\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$&$$&$\ldots$\\ |
| \end{tabular} |
\end{tabular} |
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| Those geometric series converge absolutely, |
Those geometric series converge absolutely, |
| \[ |
\[ |
| |x^k|+|x^{2k}|+|x^{3k}|+\ldots \;=\; \frac{|x|^k}{1\!-\!|x|^k} |
|x^k|+|x^{2k}|+|x^{3k}|+\ldots \;=\; \frac{|x|^k}{1\!-\!|x|^k} |
| \] |
\] |
| 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: |
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: |
| \[ |
\[ |
| \sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; x+2x^2+2x^3+3x^4+2x^5+4x^6+\ldots |
\sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; x+2x^2+2x^3+3x^4+2x^5+4x^6+\ldots |
| \] |
\] |
| Apparently, the coefficient of any $x^k$ in this power series expresses, by how many positive integers the number $k$ is \PMlinkname{divisible}{Divisibility}, i.e. the coefficient is given by the \PMlinkname{divisor function}{TauFunction} $\tau$.\, So we may write the power series form of the Lambert series as |
Apparently, the coefficient of any $x^k$ in this power series expresses, by how many positive integers the number $k$ is \PMlinkname{divisible}{Divisibility}, i.e. the coefficient is given by the \PMlinkname{divisor function}{TauFunction} $\tau$.\, So we may write the power series form of the Lambert series as |
| \[ |
\[ |
| \sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; \tau(1)x+\tau(2)x^2+\tau(3)x^3+\ldots |
\sum_{n=1}^\infty\frac{x^n}{1\!-\!x^n} \;=\; \tau(1)x+\tau(2)x^2+\tau(3)x^3+\ldots |
| \] |
\] |
