Calculus Examples

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

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

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

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Step 2.1

Substitute

$a_{n}$ and

$a_{n + 1}$ into the formula for

$r$ .

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

Step 2.2

Simplify.

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Step 2.2.1

Cancel the common factor of

${(- \frac{1}{3})}^{n + 1 - 1}$ and

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

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Step 2.2.1.1

Factor

${(- \frac{1}{3})}^{n - 1}$ out of

${(- \frac{1}{3})}^{n + 1 - 1}$ .

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

Step 2.2.1.2

Cancel the common factors.

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Step 2.2.1.2.1

Multiply by

$1$ .

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

Step 2.2.1.2.2

Cancel the common factor.

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

Step 2.2.1.2.3

Rewrite the expression.

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

Step 2.2.1.2.4

Divide

${(- \frac{1}{3})}^{n + 0 - (n - 1)}$ by

$1$ .

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

Step 2.2.2

Add

$n$ and

$0$ .

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

Step 2.2.3

Simplify each term.

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Step 2.2.3.1

Apply the distributive property.

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

Step 2.2.3.2

Multiply

$- 1$ by

$- 1$ .

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

Step 2.2.4

Subtract

$n$ from

$n$ .

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

Step 2.2.5

Add

$0$ and

$1$ .

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

Step 2.2.6

Simplify.

$r = - \frac{1}{3}$

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

$n$ into

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

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

Step 4.2

Simplify.

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Step 4.2.1

Subtract

$1$ from

$1$ .

$a = {(- \frac{1}{3})}^{0}$

Step 4.2.2

Use the power rule

${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.2.2.1

Apply the product rule to

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

$a = {(- 1)}^{0} {(\frac{1}{3})}^{0}$

Step 4.2.2.2

Apply the product rule to

$\frac{1}{3}$ .

$a = {(- 1)}^{0} \frac{1^{0}}{3^{0}}$

Step 4.2.3

Anything raised to

$0$ is

$1$ .

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

Step 4.2.4

Multiply

$\frac{1^{0}}{3^{0}}$ by

$1$ .

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

Step 4.2.5

Anything raised to

$0$ is

$1$ .

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

Step 4.2.6

Anything raised to

$0$ is

$1$ .

$a = \frac{1}{1}$

Step 4.2.7

Divide

$1$ by

$1$ .

$a = 1$

Step 5

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

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

Step 6

Simplify.

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Step 6.1

Simplify the denominator.

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Step 6.1.1

Multiply

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

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Step 6.1.1.1

Multiply

$- 1$ by

$- 1$ .

$\frac{1}{1 + 1 (\frac{1}{3})}$

Step 6.1.1.2

Multiply

$\frac{1}{3}$ by

$1$ .

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

Step 6.1.2

Write

$1$ as a fraction with a common denominator.

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

Step 6.1.3

Combine the numerators over the common denominator.

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

Step 6.1.4

Add

$3$ and

$1$ .

$\frac{1}{\frac{4}{3}}$

Step 6.2

Multiply the numerator by the reciprocal of the denominator.

$1 (\frac{3}{4})$

Step 6.3

Multiply

$\frac{3}{4}$ by

$1$ .

$\frac{3}{4}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{3}{4}$

Decimal Form:

$0.75$