sum from n=1 to infinity of 6e^{-n}

解

n = 1 \sum \infty 6 e^{- n}

6 (\frac{e}{e - 1} - 1)

十進法表記

3.49186 \dots

解答ステップ

n = 1 \sum \infty 6 e^{- n}

定数乗算の規則を適用する:

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

= 6 \cdot n = 1 \sum \infty e^{- n}

簡素化

e^{- n} : \frac{1}{e ^{n}}

= 6 \cdot n = 1 \sum \infty \frac{1}{e ^{n}}

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

= 6 (n = 0 \sum \infty \frac{1}{e ^{n}} - \frac{1}{e ^{0}})

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

= 6 (\frac{e}{e - 1} - \frac{1}{e ^{0}})

簡素化

= 6 (\frac{e}{e - 1} - 1)