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

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

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sum from n=2 to infinity of (1+1/2)^{-n}

n = 2 \sum \infty (1 + \frac{1}{2})^{- n}

\frac{4}{3}

Solution steps

n = 2 \sum \infty (1 + \frac{1}{2})^{- n}

Join

1 + \frac{1}{2} : \frac{3}{2}

= n = 2 \sum \infty (\frac{3}{2})^{- n}

Simplify

(\frac{3}{2})^{- n} : \frac{1}{( \frac{3}{2} ) ^{n}}

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

Simplify

\frac{1}{( \frac{3}{2} ) ^{n}} : (\frac{2}{3})^{n}

= n = 2 \sum \infty (\frac{2}{3})^{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{2}{3})^{n} - (\frac{2}{3})^{0} - (\frac{2}{3})^{1}

n = 0 \sum \infty (\frac{2}{3})^{n} = 3

= 3 - (\frac{2}{3})^{0} - (\frac{2}{3})^{1}

Simplify

3 - (\frac{2}{3})^{0} - (\frac{2}{3})^{1} : \frac{4}{3}

= \frac{4}{3}

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

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

n = 1 \sum \infty (\frac{1 + 5}{n})^{n}

n = 0 \sum \infty 2 (\frac{1}{3})^{n}

n = 1 \sum \infty \frac{17 n}{n !}

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