What is the sum from n=1 to infinity of 7/(n(n+3)) ?

The sum from n=1 to infinity of 7/(n(n+3)) is 77/18

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sum from n=1 to infinity of 7/(n(n+3))

n = 1 \sum \infty \frac{7}{n ( n + 3 )}

\frac{77}{18}

Solution steps

n = 1 \sum \infty \frac{7}{n ( n + 3 )}

Apply the constant multiplication rule:

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

= 7 \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 )}

= 7 \cdot n = 1 \sum \infty \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

Expand telescoping series:

\frac{11}{18}

Simplify

7 \cdot \frac{11}{18} : \frac{77}{18}

= \frac{77}{18}

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\frac{d}{d x} (1000 x - x^{2})

(\frac{x}{2})^{^{'}}

What is the sum from n=1 to infinity of 7/(n(n+3)) ?
The sum from n=1 to infinity of 7/(n(n+3)) is 77/18