What is the sum from n=1 to infinity of 5(-1/8)^{5n} ?

The sum from n=1 to infinity of 5(-1/8)^{5n} is converges

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sum from n=1 to infinity of 5(-1/8)^{5n}

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 1 \sum \infty \frac{1}{n ^{2} + 25}

n = 4 \sum \infty (- 0.61)^{n}

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

k = 1 \sum \infty \frac{k}{e ^{6 k}}

n = 1 \sum \infty \frac{n}{10 n - 3}

What is the sum from n=1 to infinity of 5(-1/8)^{5n} ?
The sum from n=1 to infinity of 5(-1/8)^{5n} is converges