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

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

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

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

1

Solution steps

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

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

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

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

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

Simplify

= 1

n = 0 \sum \infty \frac{1}{8 - n}

n = 1 \sum \infty 9 (0.1)^{n - 1}

n = 2 \sum \infty \frac{n ^{2}}{3 ^{n}}

n = 1 \sum \infty 3 (0.3)^{n - 1}

n = 1 \sum \infty \frac{5 n ^{3}}{6 ^{n}}

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