Finding the Power Series Representation for \(\ln \dfrac{1}{1-x}\)

1. Find the power series representation for \(\ln \dfrac{1}{1-x}\).
2. Find \(\ln 2\).

Solution (a) If \(y=\ln \dfrac{1}{1-x}\), then \(y^\prime = \dfrac{1}{1-x}\). That is, \(y^\prime\) is represented by the geometric series \( \sum\limits_{k\,=\,0}^{\infty }x^{k}\), which converges on the interval \( (-1,\,1)\). So, if we use the integration property of power series for \(y^\prime =\dfrac{1}{1-x}\), we obtain a series for \(y=\ln \dfrac{1}{ 1-x}\). \[ \begin{eqnarray*} y^\prime &=&\dfrac{1}{1-x}=1+x+x^{2}+\cdots +x^{n}+\cdots = \sum\limits_{k\,=\,0}^{\infty }x^{k} \\ \int_{0}^{x}\dfrac{1}{1-t}dt &=&\int_{0}^{x}( 1+t+t^{2}+\cdots +t^{n}+\cdots) ~dt \nonumber\\ \ln \dfrac{1}{1-x} &=&x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+\cdots+\dfrac{ x^{n+1}}{n+1}+\cdots\, =\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k+1}}{k+1} \end{eqnarray*} \]

The radius of convergence of this series is \(1\).

To find the interval of convergence, we investigate the endpoints. When \( x=1, \) \[ x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+\cdots+\dfrac{x^{n+1}}{n+1}+\cdots\,=1+ \dfrac{1}{2}+\dfrac{1}{3}+\cdots\, \]

the harmonic series, which diverges. When \(x=-1\), \[ x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+\cdots+\dfrac{x^{n+1}}{n+1}+\cdots\,=-1+ \dfrac{1}{2}-\dfrac{1}{3}+\cdots+(-1) ^{n+1}\dfrac{1}{n+1}+\cdots \]

an alternating harmonic series, which converges. The interval of convergence of the power series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{x^{k}}{k}\) is \( [-1,\,1) \). So, \[ \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\ln \dfrac{1}{1-x}=x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3} +\cdots\,=\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k+1}}{k+1}\qquad -1\leq x<1}} \tag{1} \end{equation*} \]

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(b) To find \(\ln 2\), notice that when \(x=-1\), we have \(\ln \dfrac{1}{1-x} =\ln \dfrac{1}{2}=- \ln 2\). In (1) let \(x=-1\). Then \[ \begin{eqnarray*} -\ln 2 &=&-1+\dfrac{1}{2}-\dfrac{1}{3}+\cdots \nonumber \\ \ln 2 &=&1-\dfrac{1}{2}+\dfrac{1}{3}\,\cdots \end{eqnarray*} \]