Power Series Expansion for Half Logarithm of 1 + x over 1 - x

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Theorem

\(\ds \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}\)
\(\ds \) \(=\) \(\ds x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots\)

valid for all $x \in \R$ such that $-1 < x < 1$.


Proof

From Power Series Expansion for $\map \ln {1 + x}$:

$(1): \quad \ds \map \ln {1 + x} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n$

for $-1 < x \le 1$.


From Power Series Expansion for $\map \ln {1 + x}$: Corollary:

$(2): \quad \ds \map \ln {1 - x} = - \sum_{n \mathop = 1}^\infty \frac {x^n} n$

for $-1 < x < 1$.


Then we have:

\(\ds \map \ln {1 + x} - \map \ln {1 - x}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n + \sum_{n \mathop = 1}^\infty \frac {x^n} n\) subtracting $(2)$ from $(1)$
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r} \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r - 1} \frac {x^{2 r} } {2 r} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r}\) separating out into odd and even indices
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} - \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r}\) $\paren {-1}^{2 r} = 1$, $\paren {-1}^{2 r - 1} = -1$
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1}\) even indices cancel
\(\ds \leadsto \ \ \) \(\ds \map \ln {\frac {1 + x} {1 - x} }\) \(=\) \(\ds 2 \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1}\) Difference of Logarithms

Hence the result.

$\blacksquare$


Sources

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