How to find the sum of this geometric serie: $ 3+ \sqrt3 + 1 + ...$

Question

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:

$S=\frac{3(3+\sqrt{3})}{4}$

Answer 1

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:

$$S=\frac{3}{1-\frac1{\sqrt3}}=\frac{3\sqrt{3}}{\sqrt3-1}=\frac{3\sqrt3(\sqrt3+1)}{2}$$

Answer 2

You also made a second error. In your working $(1+\sqrt3)(1-\sqrt3)=1-(-3)$. But actually $(1+\sqrt3)(1-\sqrt3)=1-(+3)$.

Here is a tip with geometric series: when the series converges, its sum should have the same sign as the first term and be more than half as large in absolute value; thus if the first term were $1$ the sum could be $+3/5$ but not $+2/5$ or $-3/5$.