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

The sum from k=1 to infinity of 4^{-k} is 1/3

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sum from k=1 to infinity of 4^{-k}

k = 1 \sum \infty 4^{- k}

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

k = 1 \sum \infty 4^{- k}

Simplify

4^{- k} : \frac{1}{4 ^{k}}

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

k = 0 \sum \infty \frac{1}{4 ^{k}} = \frac{4}{3}

= \frac{4}{3} - \frac{1}{4 ^{0}}

Simplify

\frac{4}{3} - \frac{1}{4 ^{0}} : \frac{1}{3}

= \frac{1}{3}

What is the sum from k=1 to infinity of 4^{-k} ?
The sum from k=1 to infinity of 4^{-k} is 1/3