What is the sum from n=1 to infinity of 9/(n(n+2)) ?

The sum from n=1 to infinity of 9/(n(n+2)) is 27/4

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sum from n=1 to infinity of 9/(n(n+2))

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

\frac{27}{4}

Solution steps

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

Apply the constant multiplication rule:

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

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

Take the partial fraction of

\frac{1}{n ( n + 2 )} : \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

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

Expand telescoping series:

\frac{3}{4}

Simplify

9 \cdot \frac{3}{4} : \frac{27}{4}

= \frac{27}{4}

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

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

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

k = 0 \sum \infty \frac{k}{k + 1}

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

What is the sum from n=1 to infinity of 9/(n(n+2)) ?
The sum from n=1 to infinity of 9/(n(n+2)) is 27/4