What is the sum from n=1 to infinity of 9^{-n} ?

The sum from n=1 to infinity of 9^{-n} is 1/8

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sum from n=1 to infinity of 9^{-n}

n = 1 \sum \infty 9^{- n}

\frac{1}{8}

Solution steps

n = 1 \sum \infty 9^{- n}

Simplify

9^{- n} : \frac{1}{9 ^{n}}

= n = 1 \sum \infty \frac{1}{9 ^{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}{9 ^{n}} - \frac{1}{9 ^{0}}

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

= \frac{9}{8} - \frac{1}{9 ^{0}}

Simplify

\frac{9}{8} - \frac{1}{9 ^{0}} : \frac{1}{8}

= \frac{1}{8}

n = 2 \sum \infty (\frac{1}{4})^{n}

k = 2 \sum \infty \frac{1}{e ^{k}}

n = 0 \sum \infty \frac{3 ^{n}}{e ^{n}}

n = 1 \sum \infty 3 n^{- 1}

k = 2 \sum \infty (- 0.75)^{k}

What is the sum from n=1 to infinity of 9^{-n} ?
The sum from n=1 to infinity of 9^{-n} is 1/8