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

解

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

\frac{1}{6}

十進法表記

0.16666 \dots

解答ステップ

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}

簡素化

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

= \frac{1}{6}