What is the sum from n=0 to infinity of 3^{-n} ?

The sum from n=0 to infinity of 3^{-n} is 3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of 3^{-n}

n = 0 \sum \infty 3^{- n}

\frac{3}{2}

Solution steps

n = 0 \sum \infty 3^{- n}

Simplify

3^{- n} : \frac{1}{3 ^{n}}

= n = 0 \sum \infty \frac{1}{3 ^{n}}

Apply Series Geometric Test:

\frac{3}{2}

= \frac{3}{2}

x \to - 4 - lim (\frac{- 5 x}{x + 4})

\frac{d y}{d x} + \frac{1}{x} y = x^{2} y^{3}

\frac{\partial}{\partial x} (\frac{1}{4 - 3 x})

x \to 0 lim (\frac{( 3 ^{x} - 5 ^{x} )}{x})

\frac{d}{d x} ((ln (2 x))^{4})

What is the sum from n=0 to infinity of 3^{-n} ?
The sum from n=0 to infinity of 3^{-n} is 3/2