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

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

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

- 2

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Geometric Test:

2

= - 2

j = 1 \sum \infty 2^{- 4 j}

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

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

n = 0 \sum \infty e^{\frac{- n ^{6}}{8}}

k = 2 \sum \infty 3 e^{- 2 k}

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