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

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

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sum from x=1 to infinity of (1/2)^{x-1}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

k = 3 \sum \infty \frac{9}{k ln ^{2} ( k )}

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

n = 0 \sum \infty (\frac{8}{5})^{n}

k = 2 \sum \infty (- 0.5)^{k}

n = 3 \sum \infty \frac{1}{n - 1}

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