What is the sum from n=2 to infinity of (1/3)^n ?

The sum from n=2 to infinity of (1/3)^n is 1/6

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sum from n=2 to infinity of (1/3)^n

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

\frac{1}{6}

Solution steps

n = 2 \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}

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

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

Simplify

\frac{3}{2} - (\frac{1}{3})^{0} - (\frac{1}{3})^{1} : \frac{1}{6}

= \frac{1}{6}

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What is the sum from n=2 to infinity of (1/3)^n ?
The sum from n=2 to infinity of (1/3)^n is 1/6