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}

Decimal

0.14285 \dots

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}

d er i v a t i v e f (x) = x^{2} + 6 x

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

d er i v a t i v e e^{c o t (4 x)}

d er i v a t i v e \frac{6 x ^{2} + 8 x + 8 x}{x}

y^{^{'}} = \frac{( y ^{2} + 2 x y )}{x ^{2}}

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