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{d}{d x} (- \frac{5}{2} x^{9})

\int_{0}^{x} ln (x) d x

\frac{d ^{2}}{d x ^{2}} (x^{3} + 3 x^{2} - 2 x + 8)

\frac{d y}{d x} = \frac{x}{y}, y (1) = 2

\frac{24}{x ^{5}}

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