Answer
converges, $\quad$ sum = $\displaystyle \frac{a}{1-b^{2}}$
Work Step by Step
An infinite geometric series is a series of the form
$ a+ar+ar^{2}+ar^{3}+\cdots+ar^{n-1}+\cdots$
An infinite geometric series for which $|r| < 1$
has the sum $S=\displaystyle \frac{a}{1-r}$
If $|r| \geq 1$, the series diverges (the sum does not exist).
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first term = $a$
common ratio: $r=b^{2}$
The square of a number that is less than 1 in magnitude,
is also less than 1 in magnitude.
So, $|r| < 1$, the series converges,
$S=\displaystyle \frac{a}{1-r} =\displaystyle \frac{a}{1-b^{2}}$
