What is the sum from k=2 to infinity of (4/5)^{3k} ?

The sum from k=2 to infinity of (4/5)^{3k} is converges

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sum from k=2 to infinity of (4/5)^{3k}

k = 2 \sum \infty (\frac{4}{5})^{3 k}

converges

Solution steps

k = 2 \sum \infty (\frac{4}{5})^{3 k}

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty 12 (- \frac{5}{8})^{n}

n = 2 \sum \infty 7 (- \frac{3}{8})^{n}

n = 2 \sum \infty (\frac{3}{5})^{4 n}

n = 0 \sum \infty 8 (\frac{1}{10})^{n}

What is the sum from k=2 to infinity of (4/5)^{3k} ?
The sum from k=2 to infinity of (4/5)^{3k} is converges