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

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

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

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

\frac{3}{2}

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \cdot k = 2 \sum \infty \frac{1}{k ^{2} - 1}

Take the partial fraction of

\frac{1}{k ^{2} - 1} : - \frac{1}{2 ( k + 1 )} + \frac{1}{2 ( k - 1 )}

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

Expand telescoping series:

\frac{3}{4}

Simplify

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

= \frac{3}{2}

y^{^{'}} + 2 y = e^{- t}, y (0) = \frac{3}{2}

3

\frac{d}{d x} (x^{7} x^{2} - 9 x + 9)

\frac{d}{d y} ((2 - x) (y - 2) (3 - x - y))

\int \frac{x + 1}{2 x + 1} d x

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