What is the sum from n=1 to infinity of (-2/3)^n ?

The sum from n=1 to infinity of (-2/3)^n is -2/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity of (-2/3)^n

n = 1 \sum \infty (- \frac{2}{3})^{n}

- \frac{2}{5}

Solution steps

n = 1 \sum \infty (- \frac{2}{3})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (- \frac{2}{3})^{n} - (- \frac{2}{3})^{0}

n = 0 \sum \infty (- \frac{2}{3})^{n} = \frac{3}{5}

= \frac{3}{5} - (- \frac{2}{3})^{0}

Simplify

\frac{3}{5} - (- \frac{2}{3})^{0} : - \frac{2}{5}

= - \frac{2}{5}

\frac{d}{d x} (ln (sec (2 x)))

f (x) = \frac{4 π ^{2}}{6 - x}

\frac{d}{d x} ((x + x^{- 1})^{3})

\frac{\partial}{\partial x} (e^{8 x + 2 y})

\frac{d}{d x} (\frac{4 x}{( 2 x + 1 ) ^{2}})

What is the sum from n=1 to infinity of (-2/3)^n ?
The sum from n=1 to infinity of (-2/3)^n is -2/5