What is the sum from n=0 to infinity of 2/(n^3-n) ?

The sum from n=0 to infinity of 2/(n^3-n) is 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

Take the partial fraction of

\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 )}

Apply the Sum Rule:

\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})

Simplify

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

= \frac{1}{2}

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

k = 0 \sum \infty (\frac{1}{7})^{k}

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

n = 0 \sum \infty (- 1)^{n} 5^{n}

n = 1 \sum \infty \frac{1}{6 n + 1}

What is the sum from n=0 to infinity of 2/(n^3-n) ?
The sum from n=0 to infinity of 2/(n^3-n) is 1/2