## Calculus: Early Transcendentals 8th Edition

$\sum ^{\infty }_{n=1}3^{n+1}4^{-n}=3\sum ^{\infty }_{n=1}\left( \dfrac {3}{4}\right) ^{n}\Rightarrow a_{1}=3\times \left( \dfrac {3}{4}\right) ^{1}=\dfrac {9}{4};r=\dfrac {3}{4} < 1\Rightarrow$ series converges $S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {9}{4}}{1-\dfrac {3}{4}}=9$