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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty e^{- 17 n}

n = 1 \sum \infty 3^{1 - n}

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

n = 1 \sum \infty \frac{1}{4 ^{n}} \cdot 2^{n}

n = 1 \sum \infty \frac{4}{2 ^{n + 2}}

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