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

解

n = 1 \sum \infty 4 (\frac{π}{4})^{n}

4 (\frac{4}{4 - π} - 1)

十進法表記

14.63916 \dots

解答ステップ

n = 1 \sum \infty 4 (\frac{π}{4})^{n}

定数乗算の規則を適用する:

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

= 4 \cdot n = 1 \sum \infty (\frac{π}{4})^{n}

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

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

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

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

簡素化

= 4 (\frac{4}{4 - π} - 1)