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

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

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

n = 0 \sum \infty \frac{2}{n ^{2} - 1}

\frac{3}{2}

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot 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 )}

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

Expand telescoping series:

\frac{3}{4}

Simplify

2 \cdot \frac{3}{4} : \frac{3}{2}

= \frac{3}{2}

\int_{0}^{\infty} f (a x) d x

\frac{d}{d y} (cos (x + y))

\frac{\partial}{\partial y} (x e^{- 2 x^{2} - 2 y^{2}})

\int sin^{2} (y) d y

y = 7 x + 4, (3, 5)

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