What is the sum from n=3 to infinity of 1/(3^n) ?

The sum from n=3 to infinity of 1/(3^n) is 1/18

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sum from n=3 to infinity of 1/(3^n)

n = 3 \sum \infty \frac{1}{3 ^{n}}

\frac{1}{18}

Solution steps

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

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

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

Simplify

\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}} - \frac{1}{3 ^{2}} : \frac{1}{18}

= \frac{1}{18}

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What is the sum from n=3 to infinity of 1/(3^n) ?
The sum from n=3 to infinity of 1/(3^n) is 1/18