## Calculus: Early Transcendentals 8th Edition

The given series can be made into a similar form as the definition of the geometric series , putting $n=1$ to get the initial term $a$ and see the multiplying factor when inputting $n=2$. Note that the summation of an infinite geometric series is $S= \frac{a}{1-r}$ if |$r$| $\lt 1$ , here $r$ is the common ratio of the geometric series. Now, If we choose n = 1 we can see that the initial term is $a$ = $4$ and the common ratio $r$ = $\frac{4}{3}$ (Just by dividing any latter term by its former, like in our case, $\frac{a_{2}}{a_{1}}$ = $\frac{\frac{48}{9}}{4}$ . |$r$| = $\frac{4}{3} \gt 1$ Thus, this geometric series diverges.