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{d y}{d x} - 4 x y = 0

\frac{\partial}{\partial y} (x^{3} + y^{2})

x \to 0 lim ((1 + a x)^{\frac{a}{x}})

x = 3 - 3 y^{2}, x = 3 y^{2} - 3

\int \frac{2 x}{( x - 5 ) ^{2}} d 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