What is the sum from n=0 to infinity of 1/(n^2-1) ?

The sum from n=0 to infinity of 1/(n^2-1) is 3/4

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sum from n=0 to infinity of 1/(n^2-1)

n = 0 \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}

k = 1 \sum \infty k e^{- k}

n = 1 \sum \infty \frac{n !}{n \cdot 2 ^{n}}

n = 1 \sum \infty \frac{1}{25 n ^{2} + 5 n - 6}

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

n = 1 \sum \infty 2 (- \frac{4}{3})^{n}

What is the sum from n=0 to infinity of 1/(n^2-1) ?
The sum from n=0 to infinity of 1/(n^2-1) is 3/4