What is the sum from n=0 to infinity of-8(-2/3)^n ?

The sum from n=0 to infinity of-8(-2/3)^n is -24/5

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sum from n=0 to infinity of-8(-2/3)^n

n = 0 \sum \infty - 8 (- \frac{2}{3})^{n}

- \frac{24}{5}

Solution steps

n = 0 \sum \infty - 8 (- \frac{2}{3})^{n}

Apply the constant multiplication rule:

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

= - 8 \cdot n = 0 \sum \infty (- \frac{2}{3})^{n}

Apply Series Geometric Test:

\frac{3}{5}

= - 8 \cdot \frac{3}{5}

Simplify

- 8 \cdot \frac{3}{5} : - \frac{24}{5}

= - \frac{24}{5}

n = 1 \sum \infty \frac{1}{3 ^{n + 1}}

n = 1 \sum \infty \frac{1}{2 n}

x = 0 \sum \infty \frac{1}{5 ^{x}}

n = 1 \sum \infty \frac{1}{1. 1 ^{n}}

k = 1 \sum \infty \frac{1}{16 k ^{2} - 4}

What is the sum from n=0 to infinity of-8(-2/3)^n ?
The sum from n=0 to infinity of-8(-2/3)^n is -24/5