Evaluating the sum of geometric series 
Possible Duplicate:
Value of $\sum\limits_n x^n$ 

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.
 A: It is indeed a geometric series:
$$
\sum_{n=0}^\infty \frac{18}{3^n}=18\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n.
$$
Can you take it from here?
A: 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
$$S=\frac{18}{1-\frac13}=\frac{18}{2/3}=27\;.$$
A: $$\sum^{\infty}_{n=0} \frac{18}{3^n} = 18 \sum^{\infty}_{n=0} \frac{1}{3^n} = \frac{18}{1-1/3} = 27$$
Where the last equality follows from the fact that $\sum^{\infty}_{n=0} x^n =\frac{1}{1-x}$ if $|x| < 1$.
A: The sum of a geometric series is
$$\sum_{k=1} ^\infty ar^k = \frac{a}{1-r}$$
where $r<1$, as the Wikipedia article says.
now simply plug in the numbers in your case, $a=18, r=\frac{1}{3}$. and you'll get
$$ \frac{18}{1-\frac{1}{3}} =  \frac{18}{\frac{2}{3}} = 27$$
Best luck with your studies.
A: $r=\text{common ratio}=1/3$.
$a=\text{first term} = 18$.
$$
\text{sum} = \frac{a}{1-r} = \frac{18}{1-(1/3)} = 27.
$$
