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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

y^{^{''}} - y^{^{'}} - 2 y = 8, y (0) = 0, y^{^{'}} (0) = 5

x \to 3 lim (4^{x^{3}} - 8 x)

x \to π - lim (sin (x))

\frac{\partial}{\partial x} (9 - x^{2} - y^{2})

\int \frac{6 x + 2}{3 x ^{2} + 2 x} d x

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