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

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

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

n = 1 \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}

\int_{0}^{1} x^{4} e^{6 x} d x

\int - t z d t

e^{2 x} y^{^{'}} - 2 e^{2 x} y = 3 e^{4 x}

x \to 5 - lim (\frac{12}{5 - x})

\int - 4 sin (3 x) + 6 sin (8 x) d x

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