sum from n=1 to infinity of 8/(n(n+2))

解

n = 1 \sum \infty \frac{8}{n ( n + 2 )}

6

解答ステップ

n = 1 \sum \infty \frac{8}{n ( n + 2 )}

定数乗算の規則を適用する:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{n ( n + 2 )}

以下の部分分数を得る:

\frac{1}{n ( n + 2 )} : \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

= 8 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

畳み込み級数を拡張する:

\frac{3}{4}

簡素化

8 \cdot \frac{3}{4} : 6

= 6