What is the sum from n=1 to infinity of (0.5)^{-n} ?

The sum from n=1 to infinity of (0.5)^{-n} is diverges

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sum from n=1 to infinity of (0.5)^{-n}

n = 1 \sum \infty (0.5)^{- n}

diverges

Solution steps

n = 1 \sum \infty 0. 5^{- n}

Simplify

0. 5^{- n} : \frac{1}{0. 5 ^{n}}

= n = 1 \sum \infty \frac{1}{0. 5 ^{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}{0. 5 ^{n}} - \frac{1}{0. 5 ^{0}}

n = 0 \sum \infty \frac{1}{0. 5 ^{n}} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int 4 x cos (3 x) d x

\frac{\partial}{\partial u} (u^{2} + v^{2})

\int \frac{6 ln ( 4 x )}{5} d x

- 2 sin (x)

f (x) = x^{5} - 3 x + 4

What is the sum from n=1 to infinity of (0.5)^{-n} ?
The sum from n=1 to infinity of (0.5)^{-n} is diverges