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

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

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sum from k=1 to infinity of (-1/2)^k

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{2 + n ^{3}}

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

n = 1 \sum \infty \frac{1 7 ^{n}}{n !}

n = 1 \sum \infty cos (\frac{18}{n})

n = 1 \sum \infty \frac{n ^{5}}{6 ^{n}}

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