calculus - Evaluating the sum of geometric series

I'm trying to understand how to evaluate the following series:
$$\sum_{n=0}^\infty {\frac{18}{3^n}}.$$

I tried following this Wikipedia Article without much success. Mathematica outputs 27 for the sum.

If someone would be kind enough to show me some light or give me an explanation I would be grateful.

Answer

Let’s assume that the series converges, and let

$$S=\sum_{n=0}^\infty\frac{18}{3^n}=\sum_{n=0}^\infty\frac{18}{3^n}=\sum_{n=0}^\infty18\left(\frac13\right)^n=\color{blue}{18\left(\frac13\right)^0}+\color{red}{18\left(\frac13\right)^1+18\left(\frac13\right)^2+18\left(\frac13\right)^3+\ldots}\;.$$

Multiply by $\frac13$:

$$\begin{align*} \frac13S&=\frac13\left(18\left(\frac13\right)^0+18\left(\frac13\right)^1+18\left(\frac13\right)^2+18\left(\frac13\right)^3+\ldots\right)\\ &=\color{red}{18\left(\frac13\right)^1+18\left(\frac13\right)^2+18\left(\frac13\right)^3+18\left(\frac13\right)^4+\ldots}\\ &=S-\color{blue}{18\left(\frac13\right)^0}\\ &=S-18\;. \end{align*}$$

Now solve the equation $\frac13S=S-18$: $\frac23S=18$, and $S=\frac32\cdot18=27$. Similar reasoning works whenever the series converges. It’s cheating a bit, though, because justifying the assumption that $S$ exists requires being able to sum the finite series $\sum_{n=0}^m\frac{18}{3^n}$ for arbitrary $m\in\Bbb N$.

Of course once you know the general formula $$\sum_{n=0}^\infty ar^n=\frac{a}{1-r}$$ when $|r|<1$, you merely observe (as I did in the first calculation) that in the sum $\displaystyle\sum_{n=0}^\infty\frac{18}{3^n}$ the terms have the form $18\left(\dfrac13\right)^n$, so $a=18$ and $r=\dfrac13$, and the formula yields

$$S=\frac{18}{1-\frac13}=\frac{18}{2/3}=27\;.$$