What is the sum from n=1 to infinity of (-1)^n2^{-n} ?

The sum from n=1 to infinity of (-1)^n2^{-n} is converges

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sum from n=1 to infinity of (-1)^n2^{-n}

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

y = sin (2 x), at x = \frac{π}{4}

\frac{\partial}{\partial x} (sin (x^{2}))

2 x - 5

\frac{d}{d x} (ln (\frac{1}{x x + 1}))

\frac{d}{d x} (arccos (\frac{1}{x}))

What is the sum from n=1 to infinity of (-1)^n2^{-n} ?
The sum from n=1 to infinity of (-1)^n2^{-n} is converges