# Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises - Page 615: 71

convergent; $S_{\infty}=2$

#### Work Step by Step

RECALL: (1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula: $S_{\infty}=\dfrac{a}{1-r}$ (2) An infinite geometric series is divergent if $|r|\ge1$. (3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it: $r = \dfrac{a_n}{a_{n-1}}$ Solve for $r$ to obtain: $\require{cancel} r = \dfrac{-\frac{3}{2}}{3} \\r=-\dfrac{3}{2} \cdot \dfrac{1}{3} \\r=-\dfrac{\cancel{3}}{2} \cdot \dfrac{1}{\cancel{3}} \\r=-\dfrac{1}{2}$ Since $|-\frac{1}{2}|<1$, then the series is convergent. Solve for the sum using the formula above, with $a=3$ and $r=-\frac{1}{2}$, to obtain: $\require{cancel} S_{\infty}=\dfrac{3}{1-(-\frac{1}{2})} \\S_{\infty}=\dfrac{3}{\frac{2}{2}+\frac{1}{2}} \\S_{\infty}=\dfrac{3}{\frac{3}{2}} \\S_{\infty} = 3 \cdot \frac{2}{3} \\S_{\infty} = \cancel{3} \cdot \dfrac{2}{\cancel{3}} \\S_{\infty}=2$

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