Answer
We show that
${\tanh ^{ - 1}}x = x + \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + \cdot\cdot\cdot$,
for $\left| x \right| \lt 1$.
Work Step by Step
Recall that $\dfrac{d}{{dx}}{\tanh ^{ - 1}}x = \dfrac{1}{{1 - {x^2}}}$. So,
${\tanh ^{ - 1}}x = \smallint \dfrac{1}{{1 - {x^2}}}{\rm{d}}x + C$, ${\ \ \ }$ where $C$ is a constant of integration.
From Table 2, we know that
$\dfrac{1}{{1 - x}} = \mathop \sum \limits_{n = 0}^\infty {x^n} = 1 + x + {x^2} + {x^3} + {x^4} + \cdot\cdot\cdot$, ${\ \ \ }$ converges to $\dfrac{1}{{1 - x}}$ for $\left| x \right| \lt 1$.
Substituting ${x^2}$ for $x$ in the series above gives
$\dfrac{1}{{1 - {x^2}}} = \mathop \sum \limits_{n = 0}^\infty {x^{2n}} = 1 + {x^2} + {x^4} + {x^6} + {x^8} + \cdot\cdot\cdot$, ${\ \ \ }$ converges to $\dfrac{1}{{1 - {x^2}}}$ for $\left| {{x^2}} \right| \lt 1$, or $\left| x \right| \lt 1$.
Therefore,
${\tanh ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \smallint {x^{2n}}{\rm{d}}x + C$
${\tanh ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{1}{{2n + 1}}{x^{2n + 1}} + C$
Setting $x=0$, we get ${\tanh ^{ - 1}}0 = 0 = C$. Thus
${\tanh ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{1}{{2n + 1}}{x^{2n + 1}}$
Hence,
${\tanh ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{1}{{2n + 1}}{x^{2n + 1}} = x + \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + \cdot\cdot\cdot$
