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

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

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

n = 1 \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}

y = \frac{3 x}{e ^{x}}

e^{- 3 x}, 2

\int \frac{3 x + 7}{( x + 4 ) ^{2}} d x

x^{x^{2}}

\frac{d}{d y} (x e^{x y})

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