Precalculus Examples

2

$2$ ,

1

$1$ ,

\frac{1}{2}

$\frac{1}{2}$ ,

\frac{1}{4}

$\frac{1}{4}$

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}{2}

$\frac{1}{2}$ gives the next term. In other words,

a_{n} = a_{1} r^{n - 1}

$a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence:

r = \frac{1}{2}

$r = \frac{1}{2}$

Step 2

The sum of a series

S_{n}

$S_{n}$ is calculated using the formula

S_{n} = \frac{a (1 - r^{n})}{1 - r}

$S_{n} = \frac{a (1 - r^{n})}{1 - r}$ . For the sum of an infinite geometric series

S_{\infty}

$S_{\infty}$ , as

n

$n$ approaches

\infty

$\infty$ ,

1 - r^{n}

$1 - r^{n}$ approaches

1

$1$ . Thus,

\frac{a (1 - r^{n})}{1 - r}

$\frac{a (1 - r^{n})}{1 - r}$ approaches

\frac{a}{1 - r}

$\frac{a}{1 - r}$ .

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

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

Step 3

The values

a = 2

$a = 2$ and

r = \frac{1}{2}

$r = \frac{1}{2}$ can be put in the equation

S_{\infty}

$S_{\infty}$ .

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

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

Step 4

Simplify the equation to find

S_{\infty}

$S_{\infty}$ .

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Step 4.1

Simplify the denominator.

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Step 4.1.1

Write

1

$1$ as a fraction with a common denominator.

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

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

Step 4.1.2

Combine the numerators over the common denominator.

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

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

Step 4.1.3

Subtract

1

$1$ from

2

$2$ .

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

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

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

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

Step 4.2

Multiply the numerator by the reciprocal of the denominator.

S_{\infty} = 2 \cdot 2

$S_{\infty} = 2 \cdot 2$

Step 4.3

Multiply

2

$2$ by

2

$2$ .

S_{\infty} = 4

$S_{\infty} = 4$

S_{\infty} = 4

$S_{\infty} = 4$

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