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

\frac{d}{d x} ((x^{2} - 5 x + 3) (4 x^{3} + 10 x - 5))

f (x) = x^{2} - 11 y^{2}

f (x) = \frac{( e ^{2 x} - 1 )}{x}

2 x + 3 y

\int_{- 2}^{- 1} (x^{2} - x - 2) d x

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