What is the sum from n=2 to infinity of 4/(n^2-1) ?

The sum from n=2 to infinity of 4/(n^2-1) is 3

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sum from n=2 to infinity of 4/(n^2-1)

n = 2 \sum \infty \frac{4}{n ^{2} - 1}

3

Solution steps

n = 2 \sum \infty \frac{4}{n ^{2} - 1}

Apply the constant multiplication rule:

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

= 4 \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 )}

= 4 \cdot n = 2 \sum \infty - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

Expand telescoping series:

\frac{3}{4}

Simplify

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

= 3

f (x) = 5.2 x + 2.3

\int \frac{( e ^{2 x} )}{1 + e ^{2 x}} d x

\frac{d y}{d x} + \frac{y}{x} = 3 x^{6} y^{2}

x \to 1 + lim (7)

(1 - x^{2}) \frac{d y}{d x} - x^{2} y = (1 + x) 1 - x^{2}

What is the sum from n=2 to infinity of 4/(n^2-1) ?
The sum from n=2 to infinity of 4/(n^2-1) is 3