sum from n=1 to infinity of 9(1/10)^n

解

n = 1 \sum \infty 9 (\frac{1}{10})^{n}

1

解答ステップ

n = 1 \sum \infty 9 (\frac{1}{10})^{n}

定数乗算の規則を適用する:

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

= 9 \cdot n = 1 \sum \infty (\frac{1}{10})^{n}

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

= 9 (n = 0 \sum \infty (\frac{1}{10})^{n} - (\frac{1}{10})^{0})

n = 0 \sum \infty (\frac{1}{10})^{n} = \frac{10}{9}

= 9 (\frac{10}{9} - (\frac{1}{10})^{0})

簡素化

9 (\frac{10}{9} - (\frac{1}{10})^{0}) : 1

= 1