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

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

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

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

\frac{3}{4}

Solution steps

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

Take the partial fraction of

\frac{1}{n ^{2} - 1} : - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

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

Expand telescoping series:

\frac{3}{4}

= \frac{3}{4}

n = 1 \sum \infty \frac{n}{n + 4}

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

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

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

n = 2 \sum \infty \frac{e ^{n}}{π ^{n}}

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