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

The sum from n=1 to infinity of 1/(8^n) is 1/7

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

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

\frac{1}{7}

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

n = 0 \sum \infty \frac{1}{8 ^{n}} = \frac{8}{7}

= \frac{8}{7} - \frac{1}{8 ^{0}}

Simplify

\frac{8}{7} - \frac{1}{8 ^{0}} : \frac{1}{7}

= \frac{1}{7}

\frac{d y}{d t} = y (7 - y) - \frac{49}{4}

f (x) = x^{2} + 6 x

\int \frac{1}{1 - 16 x ^{2}} d x

\int_{0}^{1} e^{5 x} - e^{x} d x

e^{c o t (4 x)}

What is the sum from n=1 to infinity of 1/(8^n) ?
The sum from n=1 to infinity of 1/(8^n) is 1/7