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

\frac{\partial}{\partial y} (n sin (2 (m x + n y)))

\frac{30}{( 10 s + 1 ) ( s ^{2} + 4 )}

x^{x + 2}

\frac{d}{d x} (\frac{x ^{2} - 64}{x - 8})

f (x) = 3 - x

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