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

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.

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