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

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

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

\frac{32}{9}

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}

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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