Precalculus Examples

$\sum_{k = 1}^{\infty} \frac{7}{8^{k}}$

Step 1

The sum of an infinite geometric series can be found using the formula $\frac{a}{1 - r}$ where $a$ is the first term and $r$ is the ratio between successive terms.

Step 2

Find the ratio of successive terms by plugging into the formula $r = \frac{a_{k + 1}}{a_{k}}$ and simplifying.

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Step 2.1

Substitute $a_{k}$ and $a_{k + 1}$ into the formula for $r$ .

$r = \frac{\frac{7}{8^{k + 1}}}{\frac{7}{8^{k}}}$

Step 2.2

Simplify.

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Step 2.2.1

Multiply the numerator by the reciprocal of the denominator.

$r = \frac{7}{8^{k + 1}} \cdot \frac{8^{k}}{7}$

Step 2.2.2

Combine.

$r = \frac{7 \cdot 8^{k}}{8^{k + 1} \cdot 7}$

Step 2.2.3

Cancel the common factor of $7$ .

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Step 2.2.3.1

Cancel the common factor.

$r = \frac{7 \cdot 8^{k}}{8^{k + 1} \cdot 7}$

Step 2.2.3.2

Rewrite the expression.

$r = \frac{8^{k}}{8^{k + 1}}$

Step 2.2.4

Cancel the common factor of $8^{k}$ and $8^{k + 1}$ .

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Step 2.2.4.1

Factor $8^{k + 1}$ out of $8^{k}$ .

$r = \frac{8^{k + 1} \cdot 8^{k - (k + 1)}}{8^{k + 1}}$

Step 2.2.4.2

Cancel the common factors.

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Step 2.2.4.2.1

Multiply by $1$ .

$r = \frac{8^{k + 1} \cdot 8^{k - (k + 1)}}{8^{k + 1} \cdot 1}$

Step 2.2.4.2.2

Cancel the common factor.

$r = \frac{8^{k + 1} \cdot 8^{k - (k + 1)}}{8^{k + 1} \cdot 1}$

Step 2.2.4.2.3

Rewrite the expression.

$r = \frac{8^{k - (k + 1)}}{1}$

Step 2.2.4.2.4

Divide $8^{k - (k + 1)}$ by $1$ .

$r = 8^{k - (k + 1)}$

Step 2.2.5

Simplify each term.

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Step 2.2.5.1

Apply the distributive property.

$r = 8^{k - k - 1 \cdot 1}$

Step 2.2.5.2

Multiply $- 1$ by $1$ .

$r = 8^{k - k - 1}$

Step 2.2.6

Subtract $k$ from $k$ .

$r = 8^{0 - 1}$

Step 2.2.7

Subtract $1$ from $0$ .

$r = 8^{- 1}$

Step 2.2.8

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$r = \frac{1}{8}$

Step 3

Since $| r | < 1$ , the series converges.

Step 4

Find the first term in the series by substituting in the lower bound and simplifying.

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Step 4.1

Substitute $1$ for $k$ into $\frac{7}{8^{k}}$ .

$a = \frac{7}{8^{1}}$

Step 4.2

Evaluate the exponent.

$a = \frac{7}{8}$

Step 5

Substitute the values of the ratio and first term into the sum formula.

$\frac{\frac{7}{8}}{1 - \frac{1}{8}}$

Step 6

Simplify.

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Step 6.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{7}{8} \cdot \frac{1}{1 - \frac{1}{8}}$

Step 6.2

Simplify the denominator.

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Step 6.2.1

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

$\frac{7}{8} \cdot \frac{1}{\frac{8}{8} - \frac{1}{8}}$

Step 6.2.2

Combine the numerators over the common denominator.

$\frac{7}{8} \cdot \frac{1}{\frac{8 - 1}{8}}$

Step 6.2.3

Subtract $1$ from $8$ .

$\frac{7}{8} \cdot \frac{1}{\frac{7}{8}}$

Step 6.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{7}{8} (1 (\frac{8}{7}))$

Step 6.4

Multiply $\frac{8}{7}$ by $1$ .

$\frac{7}{8} \cdot \frac{8}{7}$

Step 6.5

Cancel the common factor of $7$ .

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Step 6.5.1

Cancel the common factor.

$\frac{7}{8} \cdot \frac{8}{7}$

Step 6.5.2

Rewrite the expression.

$\frac{1}{8} \cdot 8$

Step 6.6

Cancel the common factor of $8$ .

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Step 6.6.1

Cancel the common factor.

$\frac{1}{8} \cdot 8$

Step 6.6.2

Rewrite the expression.

$1$