What is the sum from n=1 to infinity of 1/((2^n)) ?

The sum from n=1 to infinity of 1/((2^n)) is 1

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sum from n=1 to infinity of 1/((2^n))

n = 1 \sum \infty \frac{1}{( 2 ^{n} )}

1

Solution steps

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

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

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

n = 0 \sum \infty cos (\frac{n}{2})

n = 0 \sum \infty (\frac{- 1}{3})^{n}

n = 0 \sum \infty \frac{8}{9 - n}

What is the sum from n=1 to infinity of 1/((2^n)) ?
The sum from n=1 to infinity of 1/((2^n)) is 1