What is the sum from n=2 to infinity of 6/((n^2-1)) ?

The sum from n=2 to infinity of 6/((n^2-1)) is 9/2

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sum from n=2 to infinity of 6/((n^2-1))

n = 2 \sum \infty \frac{6}{( n ^{2} - 1 )}

\frac{9}{2}

Solution steps

n = 2 \sum \infty \frac{6}{n ^{2} - 1}

Apply the constant multiplication rule:

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

= 6 \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 )}

= 6 \cdot n = 2 \sum \infty - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

Expand telescoping series:

\frac{3}{4}

Simplify

6 \cdot \frac{3}{4} : \frac{9}{2}

= \frac{9}{2}

\frac{d}{d x} (\frac{9 - x ^{2}}{4 x})

\frac{\partial}{\partial x} (\frac{r t}{r + t})

\frac{d}{d x} (x^{\frac{2}{3}} (x + 4)^{\frac{1}{2}})

x^{2} y = (x^{3} + y^{3}) y^{'}

\int x^{2} x^{3} + 9 d x

What is the sum from n=2 to infinity of 6/((n^2-1)) ?
The sum from n=2 to infinity of 6/((n^2-1)) is 9/2