What is the sum from n=1 to infinity of 6^{-n} ?

The sum from n=1 to infinity of 6^{-n} is 1/5

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sum from n=1 to infinity of 6^{-n}

n = 1 \sum \infty 6^{- n}

\frac{1}{5}

Solution steps

n = 1 \sum \infty 6^{- n}

Simplify

6^{- n} : \frac{1}{6 ^{n}}

= n = 1 \sum \infty \frac{1}{6 ^{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}{6 ^{n}} - \frac{1}{6 ^{0}}

n = 0 \sum \infty \frac{1}{6 ^{n}} = \frac{6}{5}

= \frac{6}{5} - \frac{1}{6 ^{0}}

Simplify

\frac{6}{5} - \frac{1}{6 ^{0}} : \frac{1}{5}

= \frac{1}{5}

k = 2 \sum \infty (- 0.45)^{k}

k = 1 \sum \infty \frac{4}{( 2 k - 1 ) ( 2 k + 1 )}

n = 0 \sum \infty ln (\frac{n}{3 n + 1})

n = 1 \sum \infty \frac{5}{n ^{\frac{9}{8}}}

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

What is the sum from n=1 to infinity of 6^{-n} ?
The sum from n=1 to infinity of 6^{-n} is 1/5