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

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

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

n = 1 \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

n = 1 \sum \infty \frac{n + 1}{2 n + 3}

n = 0 \sum \infty \frac{1}{1 + n ^{3}}

n = 1 \sum \infty \frac{1}{2 ^{n} + n}

n = 1 \sum \infty \frac{2}{5 ^{n} - 4}

n = 0 \sum \infty \frac{n}{7 n + 1}

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