sum from n=0 to infinity of 2/(n^3-n)

解

n = 0 \sum \infty \frac{2}{n ^{3} - n}

\frac{1}{2}

十進法表記

0.5

解答ステップ

n = 2 \sum \infty \frac{2}{n ^{3} - n}

定数乗算の規則を適用する:

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

= 2 \cdot n = 2 \sum \infty \frac{1}{n ^{3} - n}

以下の部分分数を得る:

\frac{1}{n ^{3} - n} : - \frac{1}{n} + \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

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

- \frac{1}{n} = - \frac{1}{2 n} - \frac{1}{2 n}

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

総和規則を適用する:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

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

n = 2 \sum \infty - \frac{1}{2 n} + \frac{1}{2 ( n + 1 )} = - \frac{1}{4}

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

= 2 (- \frac{1}{4} + \frac{1}{2})

簡素化

2 (- \frac{1}{4} + \frac{1}{2}) : \frac{1}{2}

= \frac{1}{2}