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

The sum from n=0 to infinity of 1/(n^3-n) is 1/4

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

n = 0 \sum \infty \frac{1}{n ^{3} - n}

\frac{1}{4}

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}

\int \frac{1}{x ^{3} - 4 x ^{2} + 3 x} d x

\int 4 x (x^{2} + 5)^{3} d x

h \to 0 lim (x^{2} + 3 x h)

\int \frac{y ^{2} - 16}{y} d y

f (x) = ln (1 - x^{2})

What is the sum from n=0 to infinity of 1/(n^3-n) ?
The sum from n=0 to infinity of 1/(n^3-n) is 1/4