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

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

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

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

\frac{8}{9}

Solution steps

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

Apply Series Geometric Test:

\frac{8}{9}

= \frac{8}{9}

n = 1 \sum \infty 8 ln (\frac{n}{n + 1})

n = 0 \sum \infty 5 \cdot (\frac{1}{4})^{n}

n = 0 \sum \infty \frac{n}{n !}

n = 4 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 1 \sum \infty \frac{e ^{n}}{n ^{4}}

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