What is the sum from n=2 to infinity of 5/(n^2-1) ?

The sum from n=2 to infinity of 5/(n^2-1) is 15/4

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sum from n=2 to infinity of 5/(n^2-1)

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

\frac{15}{4}

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \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 )}

= 5 \cdot n = 2 \sum \infty - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

Expand telescoping series:

\frac{3}{4}

Simplify

5 \cdot \frac{3}{4} : \frac{15}{4}

= \frac{15}{4}

n = 1 \sum \infty \frac{1}{n ^{\frac{9}{8}}}

n = 1 \sum \infty \frac{5}{n ^{e}}

n = 1 \sum \infty \frac{0 ^{n}}{n !}

n = 1 \sum \infty \frac{5 ^{n}}{7 ^{n}}

n = 0 \sum \infty \frac{2}{( 2 n + 1 ) ^{2}}

What is the sum from n=2 to infinity of 5/(n^2-1) ?
The sum from n=2 to infinity of 5/(n^2-1) is 15/4