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

k = 2 \sum \infty (- 0.4)^{k}

n = 0 \sum \infty 7 e^{- 7 n}

n = 1 \sum \infty - \frac{1}{6 ^{n}}

k = 4 \sum \infty \frac{3}{k ln ^{2} ( k )}

n = 1 \sum \infty \frac{5 n}{n ^{2} + 1}

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