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

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

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

n = 1 \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}

n = 0 \sum \infty \frac{1}{4 - n}

n = 1 \sum \infty \frac{9}{n ( n + 2 )}

n = 1 \sum \infty \frac{4}{( - 3 ) ^{n}}

n = 0 \sum \infty \frac{2}{2 ^{n}}

n = 0 \sum \infty ln (1 - n)

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