Answer
Converges and its sum is $\frac{1}{7}$ .
Work Step by Step
Since the given series is in the same form as the definition of the geometric series that can be found in page 750 definition 4 (except for the power of $r$ being n instead of $n-1$).
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 $a$ = $\frac{1}{4}$ and the common ratio $r$ = $\frac{-3}{4}$ (Dividing any latter term by its former we can get the common ratio).
$r$ = |$\frac{-3}{4}$| $\lt 1 $ , which means the series is convergent. Now we only substitute in $S= \frac{a}{1-r}$.
$S= \frac{\frac{1}{4}}{1-\frac{-3}{4}}$ = $S= \frac{\frac{1}{4}}{\frac{7}{4}}$ = $\frac{1}{7}$ .
Hence, the geometric series converges and its sum is $\frac{1}{7}$ .
