Obtain the following formula without attempting to justify the steps used in the process.
![\[ \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5312162a8d2170952510f11b12d49774_l3.png "Rendered by QuickLaTeX.com")
We start with 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-9bf65a36fb1992f96d403414f12111d0_l3.png "Rendered by QuickLaTeX.com")
Then, assuming we can manipulate the infinite sum as if it were a finite sum, we have
![\begin{align*} &&\sum_{n=0}^{\infty} x^n &= \frac{1}{1-x} \\[9pt] \implies && \sum_{n=0}^{\infty} nx^{n-1} &= \frac{1}{(1-x)^2} &(\text{differentiate both sides)} \\[9pt] \implies && \sum_{n=1}^{\infty} nx^n &= \frac{x}{(1-x)^2} &(\text{multiply by } x \text{ and } nx^n = 0 \text{ when } n =0). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f7a9b7fd98780dd707923a0c678817a6_l3.png "Rendered by QuickLaTeX.com")
Obtain the following formula without attempting to justify the steps used in the process.
We start with the geometric series
Then, assuming we can manipulate the infinite sum as if it were a finite sum, we have