What is the sum from n=3 to infinity of (4/5)^n ?

The sum from n=3 to infinity of (4/5)^n is 64/25

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sum from n=3 to infinity of (4/5)^n

n = 3 \sum \infty (\frac{4}{5})^{n}

\frac{64}{25}

Decimal

2.56

Solution steps

n = 3 \sum \infty (\frac{4}{5})^{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{4}{5})^{n} - (\frac{4}{5})^{0} - (\frac{4}{5})^{1} - (\frac{4}{5})^{2}

n = 0 \sum \infty (\frac{4}{5})^{n} = 5

= 5 - (\frac{4}{5})^{0} - (\frac{4}{5})^{1} - (\frac{4}{5})^{2}

Simplify

5 - (\frac{4}{5})^{0} - (\frac{4}{5})^{1} - (\frac{4}{5})^{2} : \frac{64}{25}

= \frac{64}{25}

What is the sum from n=3 to infinity of (4/5)^n ?
The sum from n=3 to infinity of (4/5)^n is 64/25