What is the sum from n=0 to infinity of 9^{-n} ?

The sum from n=0 to infinity of 9^{-n} is 9/8

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sum from n=0 to infinity of 9^{-n}

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

\frac{9}{8}

Solution steps

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

Simplify

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

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

Apply Series Geometric Test:

\frac{9}{8}

= \frac{9}{8}

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

n = 0 \sum \infty \frac{i}{n !}

n = 1 \sum \infty 5^{n} i

n = 1 \sum \infty 5 (- \frac{1}{8})^{5 n}

n = 1 \sum \infty \frac{1}{n ^{2} + 25}

What is the sum from n=0 to infinity of 9^{-n} ?
The sum from n=0 to infinity of 9^{-n} is 9/8