What is the sum from n=3 to infinity of 1/(2^n) ?

The sum from n=3 to infinity of 1/(2^n) is 1/4

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sum from n=3 to infinity of 1/(2^n)

n = 3 \sum \infty \frac{1}{2 ^{n}}

\frac{1}{4}

Decimal

0.25

Solution steps

n = 3 \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}} - \frac{1}{2 ^{1}} - \frac{1}{2 ^{2}}

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

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

Simplify

2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}} - \frac{1}{2 ^{2}} : \frac{1}{4}

= \frac{1}{4}

What is the sum from n=3 to infinity of 1/(2^n) ?
The sum from n=3 to infinity of 1/(2^n) is 1/4