Chapter 9 - Infinite Series - 9.3 Infinite Series - Exercises Set 9.3 - Page 621: 4

Convergent $S=\dfrac{8}{9}$

Given $$ \sum_{k=1}^{\infty} \left(\frac{2}{3}\right)^{k+2}$$ This is a geometric series with $|r|=\frac{2}{3}<1$ , so $\displaystyle \left(\frac{2}{3}\right)^{k+2}$ is convergent. To find the sum, since $a=(2/3)^3=\frac{8}{27}$, then \begin{align*} S&=\frac{a}{1-r}\\ &=\frac{\frac{8}{27}}{1-\frac{2}{3}}\\ &=\frac{8}{9} \end{align*}