What is the sum from n=3 to infinity of (2/3)^n ?

The sum from n=3 to infinity of (2/3)^n is 8/9

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sum from n=3 to infinity of (2/3)^n

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

\frac{8}{9}

Decimal

0.88888 \dots

Solution steps

n = 3 \sum \infty (\frac{2}{3})^{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 (\frac{2}{3})^{n} - (\frac{2}{3})^{0} - (\frac{2}{3})^{1} - (\frac{2}{3})^{2}

n = 0 \sum \infty (\frac{2}{3})^{n} = 3

= 3 - (\frac{2}{3})^{0} - (\frac{2}{3})^{1} - (\frac{2}{3})^{2}

Simplify

3 - (\frac{2}{3})^{0} - (\frac{2}{3})^{1} - (\frac{2}{3})^{2} : \frac{8}{9}

= \frac{8}{9}

What is the sum from n=3 to infinity of (2/3)^n ?
The sum from n=3 to infinity of (2/3)^n is 8/9