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}

x \to 4 lim ((\frac{x ^{4}}{4}) + 1)

t cos (t)

\int (x^{2} - 1) d x

y^{^{''}} - 10 y^{^{'}} + 9 y = 27

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

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