Chapter 9 - Infinite Series - 9.3 Infinite Series - Exercises Set 9.3 - Page 621: 3

Convergence $S=\dfrac{4}{7}$

Given $$ \sum_{k=1}^{\infty} \left(\frac{-3}{4}\right)^{k-1}$$ This is a geometric series with $|r|=\frac{3}{4}<1$ , so $\displaystyle \sum_{k=1}^{\infty} \left(\frac{-3}{4}\right)^{k-1}$ is a convergent series. To find sum, since $a=(-3/4)^0=1$, then \begin{align*} S&=\frac{a}{1-r}\\ &=\frac{1}{1+\frac{3}{4}}\\ &=\frac{4}{7} \end{align*}