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

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

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

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

\frac{1}{2}

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{1}{2} \cdot 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}

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

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

= \frac{1}{2} (2 - (\frac{1}{2})^{0})

Simplify

= \frac{1}{2}

y^{^{'}} = \frac{( e ^{x} )}{e ^{y}}, y (0) = 2

n = 0 \sum \infty \frac{1}{n ^{2} + 9 n + 20}

f (x) = x^{2} - 2 x + 3

y = \frac{- 8 x}{x ^{2} + 1}

\frac{4 x}{x + 3}

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