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.

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Let’s assume that the series converges, and let


Multiply by $\frac13$:


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


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?

$$\sum^{\infty}_{n=0} \frac{18}{3^n} = 18 \sum^{\infty}_{n=0} \frac{1}{3^n} = \frac{18}{1-1/3} = 27$$

Where the last equality follows from the fact that $\sum^{\infty}_{n=0} x^n =\frac{1}{1-x}$ if $|x| < 1$.

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.

$r=\text{common ratio}=1/3$.

$a=\text{first term} = 18$.

\text{sum} = \frac{a}{1-r} = \frac{18}{1-(1/3)} = 27.