What is the sum from n=1 to infinity of 8/(n(n+2)) ?

The sum from n=1 to infinity of 8/(n(n+2)) is 6

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sum from n=1 to infinity of 8/(n(n+2))

n = 1 \sum \infty \frac{8}{n ( n + 2 )}

6

Solution steps

n = 1 \sum \infty \frac{8}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 8 \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 )}

= 8 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

8 \cdot \frac{3}{4} : 6

= 6

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

n = 0 \sum \infty \frac{1}{1 - n}

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

n = 1 \sum \infty \frac{5}{4 ^{n} - 3 ^{n}}

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

What is the sum from n=1 to infinity of 8/(n(n+2)) ?
The sum from n=1 to infinity of 8/(n(n+2)) is 6