What is the sum from k=2 to infinity of (2/3)^{3k} ?

The sum from k=2 to infinity of (2/3)^{3k} is converges

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sum from k=2 to infinity of (2/3)^{3k}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty n! e^{- 4 n}

n = 1 \sum \infty \frac{8 n}{8 n + 1}

n = 0 \sum \infty \frac{1}{3 n - 1}

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

n = 1 \sum \infty \frac{1}{n ^{2} - 4}

What is the sum from k=2 to infinity of (2/3)^{3k} ?
The sum from k=2 to infinity of (2/3)^{3k} is converges