What is the sum from n=0 to infinity of 1/((2^n)) ?

The sum from n=0 to infinity of 1/((2^n)) is 2

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sum from n=0 to infinity of 1/((2^n))

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

2

Solution steps

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

Apply Series Geometric Test:

2

= 2

P - P^{2} = \frac{d P}{d t}

\frac{\partial}{\partial x} (e^{\frac{- y}{x}})

x^{2} y^{^{'}} - 2 x y = 3 y^{4}

\int \frac{x}{8} 64 + x^{2} d x

y = x^{2} + 2, (1, 3)

What is the sum from n=0 to infinity of 1/((2^n)) ?
The sum from n=0 to infinity of 1/((2^n)) is 2