What is the sum from k=1 to infinity of 5^{-k} ?

The sum from k=1 to infinity of 5^{-k} is 1/4

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sum from k=1 to infinity of 5^{-k}

k = 1 \sum \infty 5^{- k}

\frac{1}{4}

Solution steps

k = 1 \sum \infty 5^{- k}

Simplify

5^{- k} : \frac{1}{5 ^{k}}

= k = 1 \sum \infty \frac{1}{5 ^{k}}

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

= k = 0 \sum \infty \frac{1}{5 ^{k}} - \frac{1}{5 ^{0}}

k = 0 \sum \infty \frac{1}{5 ^{k}} = \frac{5}{4}

= \frac{5}{4} - \frac{1}{5 ^{0}}

Simplify

\frac{5}{4} - \frac{1}{5 ^{0}} : \frac{1}{4}

= \frac{1}{4}

n = 1 \sum \infty n (\frac{- 5}{9})^{n}

n = 1 \sum \infty \frac{2}{n ^{2} + 6 n}

n = 2 \sum \infty \frac{1}{n ^{1.1}}

n = 5 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 5 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

What is the sum from k=1 to infinity of 5^{-k} ?
The sum from k=1 to infinity of 5^{-k} is 1/4