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

3 x + y - 2 + \frac{d y}{d x} (x - 1) = 0

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

\int \frac{1}{1 + 4 x ^{2} + 4 x ^{4}} d x

x \to \frac{π}{4} lim (2 sin (3 x))

\frac{d}{d x} ((1 - x^{4})^{3})

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