What is the sum from n=1 to infinity of 8/(n(n+3)) ?

The sum from n=1 to infinity of 8/(n(n+3)) is 44/9

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sum from n=1 to infinity of 8/(n(n+3))

n = 1 \sum \infty \frac{8}{n ( n + 3 )}

\frac{44}{9}

Solution steps

n = 1 \sum \infty \frac{8}{n ( n + 3 )}

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 + 3 )}

Take the partial fraction of

\frac{1}{n ( n + 3 )} : \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

= 8 \cdot n = 1 \sum \infty \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

Expand telescoping series:

\frac{11}{18}

Simplify

8 \cdot \frac{11}{18} : \frac{44}{9}

= \frac{44}{9}

x = 1 \sum \infty \frac{1}{x ( x + 1 )}

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

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

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

n = 1 \sum \infty \frac{1}{8 n + 1}

What is the sum from n=1 to infinity of 8/(n(n+3)) ?
The sum from n=1 to infinity of 8/(n(n+3)) is 44/9