Answer
We show that
$\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = x + \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + \cdot\cdot\cdot$,
for $\left| x \right| \lt 1$.
And conclude that $\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = {\tanh ^{ - 1}}x$,
converges for $\left| x \right| \lt 1$.
Work Step by Step
From Table 2 we get
$\ln \left( {1 + x} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{x^n}}}{n} = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \lt 1$ and $x=1$.
Substituting $-x$ for $x$ in the series above, we also get
$\ln \left( {1 - x} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( { - x} \right)}^n}}}{n} = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{2n - 1}}{x^n}}}{n} = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \cdot\cdot\cdot$
Thus,
$\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \dfrac{1}{2}\left[ {\ln \left( {1 + x} \right) - \ln \left( {1 - x} \right)} \right] = \dfrac{1}{2}\left[ {2\left( {x + \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + \cdot\cdot\cdot} \right)} \right]$
Hence, $\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = x + \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + \cdot\cdot\cdot$, for $\left| x \right| \lt 1$.
From Exercise 40, we know that ${\tanh ^{ - 1}}x = x + \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + \cdot\cdot\cdot$.
Thus, we conclude that $\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = {\tanh ^{ - 1}}x$.
