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

The sum from n=2 to infinity of 1/((n^3-n)) is 1/4

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

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

\frac{1}{4}

Decimal

0.25

Solution steps

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 )}

= 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}

= 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}

= 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}

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

Simplify

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

= \frac{1}{4}

What is the sum from n=2 to infinity of 1/((n^3-n)) ?
The sum from n=2 to infinity of 1/((n^3-n)) is 1/4