What is the sum from n=0 to infinity of 6/(n^2-1) ?

The sum from n=0 to infinity of 6/(n^2-1) is 9/2

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sum from n=0 to infinity of 6/(n^2-1)

n = 0 \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}

- 5 t^{3}

\frac{y ^{^{'}}}{1 + y ^{2}} + \frac{t}{e ^{t^{2}}} = 0, y (0) = 1

\frac{d}{d x} (\frac{11}{x})

\int 2 x (x^{2} - 7)^{6} d x

What is the sum from n=0 to infinity of 6/(n^2-1) ?
The sum from n=0 to infinity of 6/(n^2-1) is 9/2