What is the sum from k=1 to infinity of (-1/3)^k ?

The sum from k=1 to infinity of (-1/3)^k is converges

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sum from k=1 to infinity of (-1/3)^k

k = 1 \sum \infty (- \frac{1}{3})^{k}

converges

Solution steps

k = 1 \sum \infty (- \frac{1}{3})^{k}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty (\frac{3}{5})^{n - 1}

n = 0 \sum \infty \frac{5}{5 - n}

n = 1 \sum \infty \frac{1}{n ^{3} + 6}

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

n = 0 \sum \infty \frac{- 4}{e ^{n}}

What is the sum from k=1 to infinity of (-1/3)^k ?
The sum from k=1 to infinity of (-1/3)^k is converges