Verify that log ((1+x) / (1-x))1/2 = ∑ x2n+1 / (2n+1)

Assume that $\log \sqrt{\frac{1+x}{1-x}}$ has a power-series representation in terms of powers of $x$ , verify that it has the form

$\log \sqrt{\frac{1+x}{1-x}} = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}$

for real $x$ with $|x| < 1$ .

We know that for $|x|< 1$ we have the following expansions

$\begin{align*} \log (1+x) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} \\[9pt] \log (1-x) &= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}. \end{align*}$

Then we have

$\begin{align*} \log \sqrt{\frac{1+x}{1-x}} &= \frac{1}{2} \left( \log (1+x) - \log (1-x) \right) \\[9pt] &= \frac{1}{2} \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \right) \\[9pt] &= \frac{1}{2} \left( 2 \cdot \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} \right) \\[9pt] &= \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}. \end{align*}$

This was the requested identity.

Stumbling Robot

Verify that log ((1+x) / (1-x))^1/2 = ∑ x²ⁿ⁺¹ / (2n+1)

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