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

The sum from n=1 to infinity of 1/(n(n+4)) is 25/48

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sum from n=1 to infinity of 1/(n(n+4))

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

\frac{25}{48}

Solution steps

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

Take the partial fraction of

\frac{1}{n ( n + 4 )} : \frac{1}{4 n} - \frac{1}{4 ( n + 4 )}

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

Expand telescoping series:

\frac{25}{48}

= \frac{25}{48}

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

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

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

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

n = 0 \sum \infty (- \frac{1}{7})^{n}

What is the sum from n=1 to infinity of 1/(n(n+4)) ?
The sum from n=1 to infinity of 1/(n(n+4)) is 25/48