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

The sum from n=1 to infinity of 3/(n(n+3)) is 11/6

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

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

\frac{11}{6}

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \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 )}

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

Expand telescoping series:

\frac{11}{18}

Simplify

3 \cdot \frac{11}{18} : \frac{11}{6}

= \frac{11}{6}

\frac{d}{d x} (\frac{3}{( 3 - x ^{8} ) ^{\frac{3}{2}}})

\frac{d}{d x} (\frac{36.5}{x ^{2}})

\frac{d}{d x} (x^{10} e^{x + 10})

\int_{0}^{1} \frac{5}{x ( 1 + x )} d x

\frac{d y}{d t} = k (80 - y) (50 - y)

What is the sum from n=1 to infinity of 3/(n(n+3)) ?
The sum from n=1 to infinity of 3/(n(n+3)) is 11/6