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The series
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(1) |
is called Lambert series. We here consider more closely only the special case
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(2) |
for the real variable $x$ .
I. Convergence
$1^\circ.$ $x = \pm1$ : 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$$ \sqrt[n]{\frac{x^n}{1\!-\!x^n}} \;=\; \frac{x}{\sqrt[n]{1\!-\!x^n}} \;\to\; x < 1 \quad \mbox{as}\;\; n \to \infty.$$
$4^\circ.$ $-1 < x < 0$ : 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 $|x| < 1$ .
II. Power series expansion
Let $|x| < 1$ . Expand the terms to geometric series:
| $\displaystyle\frac{x}{1\!-\!x}$ |
$=$ |
$x$ |
$+$ |
$x^2$ |
$+$ |
$x^3$ |
$+$ |
$x^4$ |
$+$ |
$x^5$ |
$+$ |
$x^6$ |
$+$ |
$\ldots$ |
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| $\displaystyle\frac{x^2}{1\!-\!x^2}$ |
$=$ |
$$&$$ |
$x^2$ |
$$&$$ |
$+$ |
$x^4$ |
$$&$$ |
$+$ |
$x^6$ |
$+$ |
$\ldots$ |
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| $\displaystyle\frac{x^3}{1\!-\!x^3}$ |
$=$ |
$$&$$ |
$$&$$ |
$x^3$ |
$$&$$ |
$$&$$ |
$+$ |
$x^6$ |
$+$ |
$\ldots$ |
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| $\displaystyle\frac{x^4}{1\!-\!x^4}$ |
$=$ |
$$&$$ |
$$&$$ |
$$&$$ |
$x^4$ |
$$&$$ |
$$&$$ |
$+$ |
$\ldots$ |
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| $\displaystyle\frac{x^5}{1\!-\!x^5}$ |
$=$ |
$$&$$ |
$$&$$ |
$$&$$ |
$$&$$ |
$x^5$ |
$$&$$ |
$+$ |
$\ldots$ |
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| $\displaystyle\frac{x^6}{1\!-\!x^6}$ |
$=$ |
$$&$$ |
$$&$$ |
$$&$$ |
$$&$$ |
$$&$$ |
$x^6$ |
$+$ |
$\ldots$ |
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| $\displaystyle\;\;\ldots$ |
$$&$\ldots$&$$ |
$\ldots$ |
$$&$\ldots$&$$ |
$\ldots$ |
$$&$\ldots$&$$ |
$\ldots$ |
$$&$\ldots$\\ \end{tabular} Those geometric series converge absolutely,$$ |x^k|+|x^2k|+|x^3k|+...= |x|^k1-|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#sum;_n=1^&infin#infty;x^n1-x^n = x+2x^2+2x^3+3x^4+2x^5+4x^6+-93-JG &sum#sum;_n=1^&infin#infty;x^n1-x^n = &tau#tau;(1)x+&tau#tau;(2)x^2+&tau#tau;(3)x^3+-94-JG |
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![[parent]](http://images.planetmath.org:8080/images/uparrow.png)