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

The sum from n=1 to infinity of (0.6)^n is 1.5

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

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

1.5

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

n = 0 \sum \infty 0. 6^{n} = 2.5

= 2.5 - 0. 6^{0}

Simplify

= 1.5

k = 0 \sum \infty \frac{k}{8 k + 1}

n = 2 \sum \infty [\frac{2}{5 ^{n}}]

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

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

n = 0 \sum \infty (\frac{4}{3})^{1 - n}

What is the sum from n=1 to infinity of (0.6)^n ?
The sum from n=1 to infinity of (0.6)^n is 1.5