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

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

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

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

1

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

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

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

Simplify

= 1

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\int 2 x cos (- 4 x) d x

\frac{d}{d x} (x + ln (3 x))

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