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

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

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

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

\frac{1}{2}

Solution steps

n = 2 \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}}

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

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

Simplify

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

= \frac{1}{2}

2 x^{2} + 4 y^{2}

x \to 1 + lim (ln (x^{x - 1}))

\frac{d}{d x} ((1 + x^{2})^{3})

\frac{d}{d x} (108 x)

x \to 1 lim (\frac{x}{x + 1})

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