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}

Simplify

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

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Ratio Test:

converges

= - \frac{1}{2} converges

= converges

\int cos^{5} (x) tan^{4} (x) d x

\frac{d y}{d x} (1 + e^{x}) = e^{x - y}

f (x) = e^{x}, at x = ln (2)

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

in (2 x), (0, 2)

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