# Evaluating the sum of geometric series [duplicate]

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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## marked as duplicate by Andres Caicedo, Ayman Hourieh, Henry T. Horton, rschwieb, 5PM Jan 21 '13 at 2:22

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

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

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It's a somewhat minor nit to pick, but you should be aware that about 10% of the male population is colorblind to one degree or another. Blue is generally safe, red less so. –  Rick Decker Jan 21 '13 at 15:42

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?

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

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

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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

$$\frac{18}{1-\frac{1}{3}} = \frac{18}{\frac{2}{3}} = 27$$

$r=\text{common ratio}=1/3$.
$a=\text{first term} = 18$.
$$\text{sum} = \frac{a}{1-r} = \frac{18}{1-(1/3)} = 27.$$