What is the sum from k=1 to infinity of 5(-1/6)^{5k} ?

The sum from k=1 to infinity of 5(-1/6)^{5k} is converges

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sum from k=1 to infinity of 5(-1/6)^{5k}

k = 1 \sum \infty 5 (- \frac{1}{6})^{5 k}

converges

Solution steps

k = 1 \sum \infty 5 (- \frac{1}{6})^{5 k}

Apply the constant multiplication rule:

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

= 5 \cdot k = 1 \sum \infty (- \frac{1}{6})^{5 k}

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 1 \sum \infty \frac{3}{n ^{3} + 7}

n = 1 \sum \infty (\frac{4}{7})^{n - 1}

n = 1 \sum \infty cos (\frac{8}{n})

n = 1 \sum \infty (cos (15))^{n}

n = 0 \sum \infty (\frac{π}{6})^{n}

What is the sum from k=1 to infinity of 5(-1/6)^{5k} ?
The sum from k=1 to infinity of 5(-1/6)^{5k} is converges