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'Lambert series'
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| Title of object: |
Lambert series |
| Canonical Name: |
LambertSeries |
| Type: |
Example |
| Created on: |
2009-01-26 08:53:19 |
| Modified on: |
2009-01-26 08:53:19 |
| Classification: |
msc:40A05, msc:30B10 |
Preamble:
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\usepackage{amssymb}
\usepackage{amsmath}
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%\usepackage{psfrag}
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%\usepackage{graphicx}
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\usepackage{amsthm}
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\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
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Content:
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) $x = \pm1$:\; The series is not defined.\\
2) $|x| > 1$:\; 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) $0 \leqq x < 1$:\; The series with nonnegative \PMlinkescapetext{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) $-1 < x < 0$:\; 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, even absolutely, when\, $|x| < 1$.\\
\PMlinkescapetext{\textbf{II.\; Power series expansion}}
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