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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial x} (ln (x^{2} + 2 x))

\frac{\partial}{\partial x} (- 3 y^{3} + 3 x^{2} y^{2} - y)

y^{^{'}} = 24 x^{2} e^{- y}

\int 4 - 3 (1 - x^{2})^{- 1} d x

\frac{d}{d x} y - y = 4

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