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

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

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

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

\frac{1}{6}

Solution steps

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

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

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

k = 0 \sum \infty (\frac{1}{3})^{k} = \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 k=2 to infinity of (1/3)^k ?
The sum from k=2 to infinity of (1/3)^k is 1/6