How to find the sum of this geometric serie: $ 3+ \sqrt3 + 1 + ...$ I am trying to find the sum of this geometric series but can't find it:
$ 3+ \sqrt3 + 1 + ...$
The solution I get is:
$S=\frac{3(3+\sqrt{3})}{4}$
but the answer key shows:
$S=\frac{3\sqrt{3}(\sqrt{3}+1)}{2}$
This exercice is from a book called Pre-Calculus in a Nutshell. I could solve the other geometric series but this questio has a square root and I must be making a mistake when simplifying.
Here are the steps I took to find my solution, maybe you can see where it goes wrong?
The sum of a geometric serie is
$(S) = \frac{a}{1-r}$ when |r| < 1
$3*r=\sqrt{3}$
Therefore:
$r=\frac{\sqrt{3}}{3}$
$|\frac{3}{\sqrt{3}}|<1 $ so I can use that formula
$a=3$
which gives me
$S=\frac{3}{1-\sqrt{3}}$
simplifying I get:
$S=\frac{3*(1+\sqrt{3})}{1-\sqrt{3}*(1+\sqrt{3})}$
$S=\frac{9+3\sqrt{3}}{1- (-3)}$
Simplifying more:
$S=\frac{3(3+\sqrt{3})}{4}$

Indeed the common ratio was incorrect, it should be $\frac1{\sqrt3}=\frac{\sqrt3}{3}$
\begin{align}
S &= \frac{3}{1-\frac{\sqrt3}{3}}=\frac{9}{3-\sqrt3}=\frac{9(3+\sqrt3)}{9-3} =\frac{3(3+\sqrt3)}2 = \frac{3\sqrt3(\sqrt3+1)}{2}
\end{align}
Edit: Alternative working:
$$S=\frac{3}{1-\frac1{\sqrt3}}=\frac{3\sqrt{3}}{\sqrt3-1}=\frac{3\sqrt3(\sqrt3+1)}{2}$$
