What is the sum from n=1 to infinity of (0.8)^{n-1} ?

The sum from n=1 to infinity of (0.8)^{n-1} is converges

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sum from n=1 to infinity of (0.8)^{n-1}

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

converges

Solution steps

n = 1 \sum \infty 0. 8^{n - 1}

Apply Series Ratio Test:

converges

= converges

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

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

n = 0 \sum \infty \frac{- 3}{n ^{2} - 4}

n = 0 \sum \infty \frac{1}{n ^{100}}

n = 0 \sum \infty \frac{n ^{4}}{2 ^{n}}

What is the sum from n=1 to infinity of (0.8)^{n-1} ?
The sum from n=1 to infinity of (0.8)^{n-1} is converges