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Without justification, establish the given formula

Obtain the following formula without attempting to justify the steps used in the process.

    \[ \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1} = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right). \]


Starting with the formula for the sum of a geometric series,

    \[ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \qquad |x| < 1, \]

we substitute x^2 for x to obtain,

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

(We assumed without justification in the third line that we can integrate the series term-by-term.)

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