## College Algebra (10th Edition)

series converges; $S_{\infty} = \dfrac{20}{3}$
RECALL: (1) In the infinite geometric series: $$\sum_{k=1}^{\infty}c \cdot r^{n-1}$$ $r$ is the common ratio. (2) A geometric series converges if $|r| \lt 1$. The sum of a convergent infinite geometric series is given by the formula: $S_{\infty}=\dfrac{a_1}{1-r}$ where $r$ = common ratio $a_1$ = first term $\bf\text{Solve for r}:$ Note that when a geometric series is summation notation, the expression being raised to a power is the common ratio. Thus, the common ratio of the given series is $\dfrac{1}{4}$. Since $|\frac{1}{4}|=\frac{1}{4} \lt 1$, the series converges. $\bf\text{Find the sum:}$ The first term of the series can be evaluated by substituting $1$ for $k$: $a_1 = 5\left(\dfrac{1}{4}\right)^{1-1} = 5\left(\dfrac{1}{4}\right)^0=5(1) = 5$ With $a_1=5$ and $r=\dfrac{1}{4}$, solve for the sum using the formula in part (2) above to obtain: $S_{\infty} = \dfrac{a_1}{a-r} \\S_{\infty}=\dfrac{5}{1-\frac{1}{4}} \\S_{\infty}=\dfrac{5}{\frac{4}{4}-\frac{1}{4}} \\S_{\infty}=\dfrac{5}{\frac{3}{4}} \\S_{\infty}=5 \cdot \dfrac{4}{3} \\S_{\infty}=\dfrac{20}{3}$