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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

\frac{d}{d t} (1 + t)

\int 8 x e^{x} d x

\int 5 csc^{4} (x) cot^{6} (x) d x

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

\frac{d}{d x} (x^{3} + x + 2)

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