Calculus Examples

$\sum_{n = 1}^{\infty} \frac{2^{n}}{3^{n}}$

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 $r = \frac{a_{n + 1}}{a_{n}}$ and simplifying.

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

$r = \frac{\frac{2^{n + 1}}{3^{n + 1}}}{\frac{2^{n}}{3^{n}}}$

Simplify.

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Multiply the numerator by the reciprocal of the denominator.

$r = \frac{2^{n + 1}}{3^{n + 1}} \cdot \frac{3^{n}}{2^{n}}$

Combine.

$r = \frac{2^{n + 1} \cdot 3^{n}}{3^{n + 1} \cdot 2^{n}}$

Cancel the common factor of $2^{n + 1}$ and $2^{n}$ .

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Factor $2^{n}$ out of $2^{n + 1} \cdot 3^{n}$ .

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

Cancel the common factors.

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Factor $2^{n}$ out of $3^{n + 1} \cdot 2^{n}$ .

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

Cancel the common factor.

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

Rewrite the expression.

$r = \frac{2 \cdot 3^{n}}{3^{n + 1}}$

Cancel the common factor of $3^{n}$ and $3^{n + 1}$ .

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Factor $3^{n + 1}$ out of $2 \cdot 3^{n}$ .

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

Cancel the common factors.

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Multiply by $1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2 \cdot 3^{n - (n + 1)}$ by $1$ .

$r = 2 \cdot 3^{n - (n + 1)}$

Simplify each term.

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Apply the distributive property.

$r = 2 \cdot 3^{n - n - 1 \cdot 1}$

Multiply $- 1$ by $1$ .

$r = 2 \cdot 3^{n - n - 1}$

Subtract $n$ from $n$ .

$r = 2 \cdot 3^{0 - 1}$

Subtract $1$ from $0$ .

$r = 2 \cdot 3^{- 1}$

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

$r = 2 \cdot \frac{1}{3}$

Combine $2$ and $\frac{1}{3}$ .

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

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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Substitute $1$ for $n$ into $\frac{2^{n}}{3^{n}}$ .

$a = \frac{2^{1}}{3^{1}}$

Simplify.

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Evaluate the exponent.

$a = \frac{2}{3^{1}}$

Evaluate the exponent.

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

Step 5

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

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

Step 6

Simplify.

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Multiply the numerator by the reciprocal of the denominator.

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

Simplify the denominator.

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

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

Combine the numerators over the common denominator.

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

Subtract $2$ from $3$ .

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

Multiply the numerator by the reciprocal of the denominator.

$\frac{2}{3} (1 \cdot 3)$

Cancel the common factor of $3$ .

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Factor $3$ out of $1 \cdot 3$ .

$\frac{2}{3} (3 \cdot 1)$

Cancel the common factor.

$\frac{2}{3} (3 \cdot 1)$

Rewrite the expression.

$2$