What is the sum from k=1 to infinity of 2^{-k} ?

The sum from k=1 to infinity of 2^{-k} is 1

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sum from k=1 to infinity of 2^{-k}

k = 1 \sum \infty 2^{- k}

1

Solution steps

k = 1 \sum \infty 2^{- k}

Simplify

2^{- k} : \frac{1}{2 ^{k}}

= k = 1 \sum \infty \frac{1}{2 ^{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}{2 ^{k}} - \frac{1}{2 ^{0}}

k = 0 \sum \infty \frac{1}{2 ^{k}} = 2

= 2 - \frac{1}{2 ^{0}}

Simplify

= 1

What is the sum from k=1 to infinity of 2^{-k} ?
The sum from k=1 to infinity of 2^{-k} is 1