College Algebra 7th Edition

convergent; $S_{\infty}=\dfrac{2}{3}$
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 $|r|>1$. (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: $\require{cancel} r = \dfrac{\frac{4}{25}}{\frac{2}{5}} \\r=\dfrac{4}{25} \cdot \dfrac{5}{2} \\r=\dfrac{\cancel{4}2}{\cancel{25}5} \cdot \dfrac{\cancel{5}}{\cancel{2}} \\r=\dfrac{2}{5}$ Since $|\frac{2}{5}|<1$, then the series converges. Solve for the sum using the formula above, with $a=\dfrac{2}{5}$ and $r=\frac{2}{5}$, to obtain: $\require{cancel} S_{\infty}=\dfrac{\frac{2}{5}}{1-\frac{2}{5}} \\S_{\infty}=\dfrac{\frac{2}{5}}{\frac{5}{5}-\frac{2}{5}} \\S_{\infty}=\dfrac{\frac{2}{5}}{\frac{3}{5}} \\S_{\infty} = \dfrac{2}{5} \cdot \frac{5}{3} \\S_{\infty} = \dfrac{2}{\cancel{5}} \cdot \dfrac{\cancel{5}}{3} \\S_{\infty}=\dfrac{2}{3}$