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

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

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

n = 2 \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}

\frac{d y}{d x} = 3 - y

\int \frac{x + 1}{x ^{2} - 4} d x

f (x) = - \frac{8}{x}

x \to 3 lim (- 2 (x - 2)^{2} + 4)

- \frac{1}{2} cos (2 x)

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