What is the sum from n=0 to infinity of 8^{-n} ?

The sum from n=0 to infinity of 8^{-n} is 8/7

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sum from n=0 to infinity of 8^{-n}

n = 0 \sum \infty 8^{- n}

\frac{8}{7}

Solution steps

n = 0 \sum \infty 8^{- n}

Simplify

8^{- n} : \frac{1}{8 ^{n}}

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

Apply Series Geometric Test:

\frac{8}{7}

= \frac{8}{7}

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

n = 1 \sum \infty \frac{n ^{3}}{n ^{2}}

n = 1 \sum \infty \frac{26}{n !}

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

n = 1 \sum \infty \frac{1}{2 + n ^{3}}

What is the sum from n=0 to infinity of 8^{-n} ?
The sum from n=0 to infinity of 8^{-n} is 8/7