What is the sum from n=3 to infinity of 5/(9^n) ?

The sum from n=3 to infinity of 5/(9^n) is 5/648

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sum from n=3 to infinity of 5/(9^n)

n = 3 \sum \infty \frac{5}{9 ^{n}}

\frac{5}{648}

Solution steps

n = 3 \sum \infty \frac{5}{9 ^{n}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 \cdot n = 3 \sum \infty \frac{1}{9 ^{n}}

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

= 5 (n = 0 \sum \infty \frac{1}{9 ^{n}} - \frac{1}{9 ^{0}} - \frac{1}{9 ^{1}} - \frac{1}{9 ^{2}})

n = 0 \sum \infty \frac{1}{9 ^{n}} = \frac{9}{8}

= 5 (\frac{9}{8} - \frac{1}{9 ^{0}} - \frac{1}{9 ^{1}} - \frac{1}{9 ^{2}})

Simplify

5 (\frac{9}{8} - \frac{1}{9 ^{0}} - \frac{1}{9 ^{1}} - \frac{1}{9 ^{2}}) : \frac{5}{648}

= \frac{5}{648}

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What is the sum from n=3 to infinity of 5/(9^n) ?
The sum from n=3 to infinity of 5/(9^n) is 5/648