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

The sum from n=1 to infinity of 2/(n(n+4)) is 25/24

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

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

\frac{25}{24}

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \cdot 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 )}

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

Expand telescoping series:

\frac{25}{48}

Simplify

2 \cdot \frac{25}{48} : \frac{25}{24}

= \frac{25}{24}

n = 1 \sum \infty (- 0.6)^{n}

n = 1 \sum \infty \frac{2}{n ^{3} + 5}

k = 6 \sum \infty \frac{1}{( k - 5 ) ^{3}}

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

n = 1 \sum \infty \frac{1}{n ^{2} + 7 n + 1}

What is the sum from n=1 to infinity of 2/(n(n+4)) ?
The sum from n=1 to infinity of 2/(n(n+4)) is 25/24