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

The sum from n=1 to infinity of (-2)^{-n} is converges

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)^{-n}

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

converges

Solution steps

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

Simplify

(- 2)^{- n} : \frac{1}{( - 2 ) ^{n}}

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{6 ^{n - 1}}

n = 5 \sum \infty \frac{1}{( n - 4 ) ^{3}}

k = 1 \sum \infty \frac{1}{5 ^{k} - 2 ^{k}}

k = 5 \sum \infty \frac{6}{k ln ^{2} ( k )}

n = 0 \sum \infty \frac{i}{n ^{4}}

What is the sum from n=1 to infinity of (-2)^{-n} ?
The sum from n=1 to infinity of (-2)^{-n} is converges