Precalculus Examples

$\sum_{k = 1}^{\infty} 3 {(\frac{2}{3})}^{k - 1}$

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 $r$ .

$r = \frac{3 {(\frac{2}{3})}^{k + 1 - 1}}{3 {(\frac{2}{3})}^{k - 1}}$

Step 2.2

Simplify.

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Step 2.2.1

Cancel the common factor of $3$ .

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Step 2.2.1.1

Cancel the common factor.

$r = \frac{3 {(\frac{2}{3})}^{k + 1 - 1}}{3 {(\frac{2}{3})}^{k - 1}}$

Step 2.2.1.2

Rewrite the expression.

$r = \frac{{(\frac{2}{3})}^{k + 1 - 1}}{{(\frac{2}{3})}^{k - 1}}$

Step 2.2.2

Cancel the common factor of ${(\frac{2}{3})}^{k + 1 - 1}$ and ${(\frac{2}{3})}^{k - 1}$ .

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Step 2.2.2.1

Factor ${(\frac{2}{3})}^{k - 1}$ out of ${(\frac{2}{3})}^{k + 1 - 1}$ .

$r = \frac{{(\frac{2}{3})}^{k - 1} {(\frac{2}{3})}^{k + 0 - (k - 1)}}{{(\frac{2}{3})}^{k - 1}}$

Step 2.2.2.2

Cancel the common factors.

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Step 2.2.2.2.1

Multiply by $1$ .

$r = \frac{{(\frac{2}{3})}^{k - 1} {(\frac{2}{3})}^{k + 0 - (k - 1)}}{{(\frac{2}{3})}^{k - 1} \cdot 1}$

Step 2.2.2.2.2

Cancel the common factor.

$r = \frac{{(\frac{2}{3})}^{k - 1} {(\frac{2}{3})}^{k + 0 - (k - 1)}}{{(\frac{2}{3})}^{k - 1} \cdot 1}$

Step 2.2.2.2.3

Rewrite the expression.

$r = \frac{{(\frac{2}{3})}^{k + 0 - (k - 1)}}{1}$

Step 2.2.2.2.4

Divide ${(\frac{2}{3})}^{k + 0 - (k - 1)}$ by $1$ .

$r = {(\frac{2}{3})}^{k + 0 - (k - 1)}$

Step 2.2.3

Add $k$ and $0$ .

$r = {(\frac{2}{3})}^{k - (k - 1)}$

Step 2.2.4

Simplify each term.

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Step 2.2.4.1

Apply the distributive property.

$r = {(\frac{2}{3})}^{k - k - - 1}$

Step 2.2.4.2

Multiply $- 1$ by $- 1$ .

$r = {(\frac{2}{3})}^{k - k + 1}$

Step 2.2.5

Subtract $k$ from $k$ .

$r = {(\frac{2}{3})}^{0 + 1}$

Step 2.2.6

Add $0$ and $1$ .

$r = {(\frac{2}{3})}^{1}$

Step 2.2.7

Simplify.

$r = \frac{2}{3}$

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 $3 {(\frac{2}{3})}^{k - 1}$ .

$a = 3 {(\frac{2}{3})}^{1 - 1}$

Step 4.2

Simplify.

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Step 4.2.1

Subtract $1$ from $1$ .

$a = 3 {(\frac{2}{3})}^{0}$

Step 4.2.2

Apply the product rule to $\frac{2}{3}$ .

$a = 3 \frac{2^{0}}{3^{0}}$

Step 4.2.3

Anything raised to $0$ is $1$ .

$a = 3 \frac{1}{3^{0}}$

Step 4.2.4

Anything raised to $0$ is $1$ .

$a = 3 (\frac{1}{1})$

Step 4.2.5

Cancel the common factor of $1$ .

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Step 4.2.5.1

Cancel the common factor.

$a = 3 (\frac{1}{1})$

Step 4.2.5.2

Rewrite the expression.

$a = 3 \cdot 1$

Step 4.2.6

Multiply $3$ by $1$ .

$a = 3$

Step 5

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

$\frac{3}{1 - \frac{2}{3}}$

Step 6

Simplify.

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Step 6.1

Simplify the denominator.

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Step 6.1.1

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

$\frac{3}{\frac{3}{3} - \frac{2}{3}}$

Step 6.1.2

Combine the numerators over the common denominator.

$\frac{3}{\frac{3 - 2}{3}}$

Step 6.1.3

Subtract $2$ from $3$ .

$\frac{3}{\frac{1}{3}}$

Step 6.2

Multiply the numerator by the reciprocal of the denominator.

$3 \cdot 3$

Step 6.3

Multiply $3$ by $3$ .

$9$