Precalculus Examples

$\frac{1}{3} + \frac{1}{9} + \frac{1}{27}$

Step 1

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by $\frac{1}{3}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{1}{3}$

Step 2

The sum of a series $S_{n}$ is calculated using the formula $S_{n} = \frac{a (1 - r^{n})}{1 - r}$ . For the sum of an infinite geometric series $S_{\infty}$ , as $n$ approaches $\infty$ , $1 - r^{n}$ approaches $1$ . Thus, $\frac{a (1 - r^{n})}{1 - r}$ approaches $\frac{a}{1 - r}$ .

$S_{\infty} = \frac{a}{1 - r}$

Step 3

The values $a = \frac{1}{3}$ and $r = \frac{1}{3}$ can be put in the equation $S_{\infty}$ .

$S_{\infty} = \frac{\frac{1}{3}}{1 - \frac{1}{3}}$

Step 4

Simplify the equation to find $S_{\infty}$ .

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Step 4.1

Multiply the numerator by the reciprocal of the denominator.

$S_{\infty} = \frac{1}{3} \cdot \frac{1}{1 - \frac{1}{3}}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Write $1$ as a fraction with a common denominator.

$S_{\infty} = \frac{1}{3} \cdot \frac{1}{\frac{3}{3} - \frac{1}{3}}$

Step 4.2.2

Combine the numerators over the common denominator.

$S_{\infty} = \frac{1}{3} \cdot \frac{1}{\frac{3 - 1}{3}}$

Step 4.2.3

Subtract $1$ from $3$ .

$S_{\infty} = \frac{1}{3} \cdot \frac{1}{\frac{2}{3}}$

Step 4.3

Multiply the numerator by the reciprocal of the denominator.

$S_{\infty} = \frac{1}{3} \cdot (1 (\frac{3}{2}))$

Step 4.4

Multiply $\frac{3}{2}$ by $1$ .

$S_{\infty} = \frac{1}{3} \cdot \frac{3}{2}$

Step 4.5

Cancel the common factor of $3$ .

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Step 4.5.1

Cancel the common factor.

$S_{\infty} = \frac{1}{3} \cdot \frac{3}{2}$

Step 4.5.2

Rewrite the expression.

$S_{\infty} = \frac{1}{2}$