What is the sum from k=1 to infinity of 3(-1/8)^{3k} ?

The sum from k=1 to infinity of 3(-1/8)^{3k} is converges

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sum from k=1 to infinity of 3(-1/8)^{3k}

k = 1 \sum \infty 3 (- \frac{1}{8})^{3 k}

converges

Solution steps

k = 1 \sum \infty 3 (- \frac{1}{8})^{3 k}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot k = 1 \sum \infty (- \frac{1}{8})^{3 k}

Apply Series Ratio Test:

converges

= 3 converges

= converges

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

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

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

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

n = 1 \sum \infty \frac{5 n}{n ^{2} + 9}

What is the sum from k=1 to infinity of 3(-1/8)^{3k} ?
The sum from k=1 to infinity of 3(-1/8)^{3k} is converges