What is the sum from n=1 to infinity of (0.7)^n ?

The sum from n=1 to infinity of (0.7)^n is 2.33333…

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sum from n=1 to infinity of (0.7)^n

n = 1 \sum \infty (0.7)^{n}

2.33333 \dots

Solution steps

n = 1 \sum \infty 0. 7^{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. 7^{n} - 0. 7^{0}

n = 0 \sum \infty 0. 7^{n} = 3.33333 \dots

= 3.33333 \dots - 0. 7^{0}

Simplify

= 2.33333 \dots

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

k = 0 \sum \infty (\frac{1}{19})^{k}

n = 1 \sum \infty (\frac{3}{2})^{3 - 2 n}

n = 0 \sum \infty 6 (\frac{8}{9})^{n}

n = 0 \sum \infty 0.5

What is the sum from n=1 to infinity of (0.7)^n ?
The sum from n=1 to infinity of (0.7)^n is 2.33333…