What is the sum from n=0 to infinity of 4*(-1/8)^n ?

The sum from n=0 to infinity of 4*(-1/8)^n is 32/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=0 to infinity of 4*(-1/8)^n

n = 0 \sum \infty 4 \cdot (- \frac{1}{8})^{n}

\frac{32}{9}

Decimal

3.55555 \dots

Solution steps

n = 0 \sum \infty 4 (- \frac{1}{8})^{n}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty (- \frac{1}{8})^{n}

Apply Series Geometric Test:

\frac{8}{9}

= 4 \cdot \frac{8}{9}

Simplify

4 \cdot \frac{8}{9} : \frac{32}{9}

= \frac{32}{9}

y \to - 2 lim (\frac{y ^{3} + 8}{y + 2})

\int cos (x) sec (x) tan (x) d x

\int_{0}^{2} 2 x - 1 d x

\frac{\partial}{\partial x} (lo g_{y} (x))

\frac{\partial}{\partial x} (x^{3} y + 7 x^{2} y^{2})

What is the sum from n=0 to infinity of 4*(-1/8)^n ?
The sum from n=0 to infinity of 4*(-1/8)^n is 32/9