Answer
We find common ratio $r$ by dividing any latter term by its former we can get the common ration (r). $\frac{9}{4} \div 3 = \frac{3}{4}$ . So , $r = \frac{3}{4}$
|$\frac{3}{4}|\lt1 $ , meaning that this geometric series will converge.
By the formula from definition 4 in page 750 , with $a$ being $4$ and using the calculated r and we get
$S= \frac{a}{1-r}$ = $\frac{4}{1-\frac{3}{4}} = \frac{4}{\frac{1}{4}} = 16$
Result: the series converges and its sum is 16 .
Work Step by Step
We look at the ratio change from term to term we find that (Same as exercise 17) :
Dividing any latter term by its former we can get the common ration (r). $\frac{9}{4} \div 3 = \frac{3}{4}$ . Also, $3\div 4 = \frac{3}{4}$ .
Note that |$\frac{3}{4}|\lt1 $ which means that this geometric series will converge.
Now, use the formula from definition 4 in page 750 , knowing that $a$ refers to the first term which is $4$ and using the calculated r and we get
$S= \frac{a}{1-r}$ = $\frac{4}{1-\frac{3}{4}} = \frac{4}{\frac{1}{4}} = 16$
Result: the series converges and its sum is 16 .
