Obtain the following formula without attempting to justify the steps used in the process.
![\[ \sum_{n=1}^{\infty} \frac{x^n}{n} = \log \left( \frac{1}{1-x} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c7bee8572752ea4b83bdc55aca89d2a0_l3.png "Rendered by QuickLaTeX.com")
Starting with the formula for the geometric series,
![\[ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \qquad \text{for } |x| < 1, \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6a80731de8da2106ab18807ec35b39ac_l3.png "Rendered by QuickLaTeX.com")
We integrate both sides, (and assuming without justification that we can integrate the series term-by-term)
![\begin{align*} && \sum_{n=0}^{\infty} x^n &= \frac{1}{1-x} \\[9pt] \implies && \int \sum_{n=0}^{\infty} x^n &= \int \frac{1}{1-x} \, dx \\[9pt] \implies && \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} &= \log \frac{1}{1-x} \, dx \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{x^n}{n} &= \log \frac{1}{1-x}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-695ae9aa3a05808afec402334e58768b_l3.png "Rendered by QuickLaTeX.com")