What is the sum from n=1 to infinity of (0.6)^{n-1} ?

The sum from n=1 to infinity of (0.6)^{n-1} is converges

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sum from n=1 to infinity of (0.6)^{n-1}

n = 1 \sum \infty (0.6)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 0. 6^{n - 1}

Apply Series Ratio Test:

converges

= converges

n = 3 \sum \infty \frac{1}{n ^{2} + n}

n = 0 \sum \infty \frac{1}{n ^{0.9}}

n = 0 \sum \infty π

n = 1 \sum \infty (- 1)^{2 n + 1}

n = 1 \sum \infty (\frac{1}{5})^{n} + \frac{3}{n}

What is the sum from n=1 to infinity of (0.6)^{n-1} ?
The sum from n=1 to infinity of (0.6)^{n-1} is converges