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

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

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

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

1

Solution steps

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

Simplify

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

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

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

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

Simplify

= 1

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