What is the sum from n=0 to infinity of 1/(2^{3n)} ?

The sum from n=0 to infinity of 1/(2^{3n)} is 8/7

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sum from n=0 to infinity of 1/(2^{3n)}

n = 0 \sum \infty \frac{1}{2 ^{3 n}}

\frac{8}{7}

Solution steps

n = 0 \sum \infty \frac{1}{2 ^{3 n}}

2^{3 n} = 8^{n}

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

Apply Series Geometric Test:

\frac{8}{7}

= \frac{8}{7}

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

k = 1 \sum \infty \frac{k}{ln ( k )}

n = 1 \sum \infty \frac{7 ^{n}}{n ! n}

n = 0 \sum \infty 8 (\frac{1}{9})^{n}

n = 0 \sum \infty e^{- 18 n}

What is the sum from n=0 to infinity of 1/(2^{3n)} ?
The sum from n=0 to infinity of 1/(2^{3n)} is 8/7