Summation of a convergent series I have the following problem:

So I start as follows:
$B_{2}M_{2}=\frac{1}{\sqrt{3}}$ and I realize that $B_{2}M_{2} = A_{2}B_{2}$, so $B_{3}M_{3} = \frac{1}{\sqrt{3}}^{2}$. Next, I compute $A_{1}B_{2}=A_{1}B_{1}-B_{2}M_{2} = 1-\frac{1}{\sqrt{3}}$. For $A_{2}B_{3}$, we can do something similar, namely: $A_{2}B_{3} = A_{2}B_{2}-B_{3}M_{3}=\frac{1}{\sqrt{3}}-(\frac{1}{\sqrt{3}})^{2}$.
We can now derive the formula for computing the area of both triangles:
$$\frac{1}{2}\sqrt{(A_{1}B_{1})^{2}+(B_{1}M_{1})^{2}}\cdot A_{1}B_{2} \sin(30)$$
Now since $A_{1}B_{1}=B_{1}M_{1}$, we can simplify it to:
$$\frac{1}{\sqrt{2}}\sqrt{(A_{1}B_{1})^{2}}\cdot A_{1}B_{2}$$
This can be written as:
$$S_{n} = \frac{1}{\sqrt{2}}\sqrt{(A_{n}B_{n})^{2}}\cdot A_{n}B_{n+1}$$
I fill in the missing pieces and I get:
$$S_{n} = \frac{1}{\sqrt{2}} \cdot \sqrt{\left(\frac{1}{3}\right)^{n-1}} \cdot \left( \left(\frac{1}{3} \right) ^{n-1} - \left( \frac{1}{3} \right)^{n} \right) $$
Now I simplify it to:
$$S_n = \sqrt{6} \cdot 3^{-1.5n}$$
Now I have a small problem, if I try to solve $\sum_{n=1}^{\infty} \sqrt{6} \cdot 3^{-1.5n}$, I know we can rewrite it to $\sqrt{6} \sum_{n=1}^{\infty} 3^{-1.5n}$, but I have no idea how to continue to get the closed form.
In addition, I was wondering, is there a faster way to get the solution? 

Hint. One may recall the evaluation of a geometric series
$$
\sum_{n=1}^\infty r^n=\frac{r}{1-r}, \quad |r|<1.
$$ Then put $r=3^{-1.5}=0.1924\cdots$ to get the answer.
