Evaluate the Summation sum from x=1 to infinity of (-2/3)^(x-1)

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

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.

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

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

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

Simplify.

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

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

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

Cancel the common factors.

Multiply by $1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Add $x$ and $0$ .

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 1$ by $- 1$ .

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

Subtract $x$ from $x$ .

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

Add $0$ and $1$ .

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

Simplify.

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

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

Substitute $1$ for $x$ into ${(- \frac{2}{3})}^{x - 1}$ .

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

Simplify.

Subtract $1$ from $1$ .

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

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

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

$a = {(- 1)}^{0} {(\frac{2}{3})}^{0}$

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

$a = {(- 1)}^{0} \frac{2^{0}}{3^{0}}$

Anything raised to $0$ is $1$ .

$a = 1 \frac{2^{0}}{3^{0}}$

Multiply $\frac{2^{0}}{3^{0}}$ by $1$ .

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

Anything raised to $0$ is $1$ .

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

Anything raised to $0$ is $1$ .

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

Divide $1$ by $1$ .

$a = 1$

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

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

Simplify.

Simplify the denominator.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{2}{3}$ by $1$ .

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

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

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

Combine the numerators over the common denominator.

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

Add $3$ and $2$ .

$\frac{1}{\frac{5}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$1 (\frac{3}{5})$

Multiply $\frac{3}{5}$ by $1$ .

$\frac{3}{5}$

The result can be shown in multiple forms.

Exact Form:

$\frac{3}{5}$

Decimal Form:

$0.6$

Evaluate the Summation sum from x=1 to infinity of (-2/3)^(x-1)

Related Questions

Support LCMGCF.com by disabling your adblocker.