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}

\frac{d}{d x} (ln (3 x^{2} + 1))

x \to 1 - lim (- \frac{2}{1 - x})

x \to 0 lim (ln (sin (x)))

x, y = sin (x y)

(ln (2 x))^{2}

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