What is the sum from n=1 to infinity of (-7/8)^{n-1} ?

The sum from n=1 to infinity of (-7/8)^{n-1} is converges

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sum from n=1 to infinity of (-7/8)^{n-1}

n = 1 \sum \infty (- \frac{7}{8})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (- \frac{7}{8})^{n - 1}

Apply Series Ratio Test:

converges

= converges

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

n = 0 \sum \infty (- 1)^{n - 1}

n = 1 \sum \infty (9.9)^{n}

n = 1 \sum \infty 7 (0.8)^{n - 1}

n = 1 \sum \infty \frac{n}{13 n + 4}

What is the sum from n=1 to infinity of (-7/8)^{n-1} ?
The sum from n=1 to infinity of (-7/8)^{n-1} is converges