Answer
converges, $\quad $sum = $\displaystyle \frac{1}{2}(3+\sqrt{3})$
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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$a=1$
$ r=\displaystyle \frac{3^{-1/2}}{1}=3^{-1/2}=\frac{1}{\sqrt{3}}\approx 0.5774, \quad$
$ |r| < 1, $so the series is convergent,
$S=\displaystyle \frac{1}{1-\frac{1}{\sqrt{3}}}= \frac{1}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}}{\sqrt{3}-1}$
$=\displaystyle \frac{\sqrt{3}}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{3+\sqrt{3}}{3-1}$
$=\displaystyle \frac{1}{2}(3+\sqrt{3})$
