What is the sum from k=1 to infinity of (-1/4)^k ?

The sum from k=1 to infinity of (-1/4)^k is converges

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sum from k=1 to infinity of (-1/4)^k

k = 1 \sum \infty (- \frac{1}{4})^{k}

converges

Solution steps

k = 1 \sum \infty (- \frac{1}{4})^{k}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{n ^{4}}{3 ^{n}}

n = 0 \sum \infty \frac{7}{n ^{2} + 9}

n = 1 \sum \infty \frac{9}{n ^{2} + n ^{3}}

k = 3 \sum \infty \frac{7}{k ln ^{2} ( k )}

k = 2 \sum \infty \frac{k}{ln ( 5 k )}

What is the sum from k=1 to infinity of (-1/4)^k ?
The sum from k=1 to infinity of (-1/4)^k is converges