What is the sum from n=0 to infinity of 6/(n(n+2)) ?

The sum from n=0 to infinity of 6/(n(n+2)) is 9/2

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sum from n=0 to infinity of 6/(n(n+2))

n = 0 \sum \infty \frac{6}{n ( n + 2 )}

\frac{9}{2}

Solution steps

n = 1 \sum \infty \frac{6}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 6 \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 )}

= 6 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

6 \cdot \frac{3}{4} : \frac{9}{2}

= \frac{9}{2}

k = 1 \sum \infty \frac{k}{k ^{2} - 1}

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

n = 0 \sum \infty 1 - e^{- n}

k = 1 \sum \infty \frac{k}{k ^{2} + 4}

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

What is the sum from n=0 to infinity of 6/(n(n+2)) ?
The sum from n=0 to infinity of 6/(n(n+2)) is 9/2