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

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

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

n = 2 \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}

\frac{d y}{d t} = 0.2 \cdot (y - sin (t)) + cos (t)

\int \frac{1}{3 x + 5} d x

\int (x + 1) (x + 4) d x

\int \frac{1}{x x ^{2} - a ^{2}} d x

x \to 6 lim (2 x)

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