sum from n=1 to infinity of (e^n)/(4^n)

解

n = 1 \sum \infty \frac{e ^{n}}{4 ^{n}}

\frac{4}{4 - e} - 1

十進法表記

2.12081 \dots

解答ステップ

n = 1 \sum \infty \frac{e ^{n}}{4 ^{n}}

指数の規則を適用する:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= n = 1 \sum \infty (\frac{e}{4})^{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{e}{4})^{n} - (\frac{e}{4})^{0}

n = 0 \sum \infty (\frac{e}{4})^{n} = \frac{4}{4 - e}

= \frac{4}{4 - e} - (\frac{e}{4})^{0}

簡素化

= \frac{4}{4 - e} - 1