What is the sum from k=0 to infinity of (-1/2)^k ?

The sum from k=0 to infinity of (-1/2)^k is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=0 to infinity of (-1/2)^k

k = 0 \sum \infty (- \frac{1}{2})^{k}

converges

Solution steps

k = 0 \sum \infty (- \frac{1}{2})^{k}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{4 n + 4}

n = n = 1 \sum \infty \frac{1}{2 ^{n}}

k = 0 \sum \infty (\frac{3}{4})^{k}

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

n = 6 \sum \infty \frac{2}{n - 5}

What is the sum from k=0 to infinity of (-1/2)^k ?
The sum from k=0 to infinity of (-1/2)^k is converges