<s>[INST] Algebraic Symbol Manipulation While Finding the Sum of a Series In a precalc text, in the chapter on geometric progressions and series, we're told of the formula
$S=\frac{a(1-r^{n+1})}{1-r}$
and that: 
$S=\frac{a}{1-r}$ is valid for $|r|<1$
We're then asked to find the sum of $3+\sqrt{3}+1+...$
Literally that's all the information given, including the "...".
The text gives the answer as: $\frac{3\sqrt{3}(\sqrt{3}+1)}{2}$ but I'm struggling to derive the calculations that get us there.
I set $a=3$ and $r = \frac{\sqrt{3}}{3}$ which, by the 2nd formula above, gives me $\frac{3}{1-\frac{\sqrt{3}}{3}}$
I clear the radical sign in the denominator by multiplying by the conjugate:
$\frac{3}{1-\frac{\sqrt{3}}{3}}\cdot\frac{1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}$ which gives me $\frac{3+\frac{3\sqrt{3}}{3}}{\frac{2}{3}}\Rightarrow3+\frac{3\sqrt{3}}{3}\cdot\frac{3}{2}\Rightarrow...?$ 
It's at this point the head scratching begins.
Hints?
 [/INST] Hint:
$$\begin{align}
\frac{3}{1-\frac{1}{\sqrt{3}}}&=\frac{3}{1-\frac{1}{\sqrt{3}}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\
&=\frac{3\sqrt{3}}{\sqrt{3}-1}\\
&=\frac{3\sqrt{3}}{\sqrt{3}-1}\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1}
\end{align}$$
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