What is the sum from k=1 to infinity of (-6/7)^k ?

The sum from k=1 to infinity of (-6/7)^k is converges

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sum from k=1 to infinity of (-6/7)^k

k = 1 \sum \infty (- \frac{6}{7})^{k}

converges

Solution steps

k = 1 \sum \infty (- \frac{6}{7})^{k}

Apply Series Ratio Test:

converges

= converges

k = 1 \sum \infty (\frac{2}{3})^{k + 2}

n = 1 \sum \infty \frac{4}{n ! + 4 ^{n}}

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

n = 2 \sum \infty (- 0.8)^{n}

n = 0 \sum \infty 7 (\frac{3}{4})^{n}

What is the sum from k=1 to infinity of (-6/7)^k ?
The sum from k=1 to infinity of (-6/7)^k is converges