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

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

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

n = 1 \sum \infty \frac{6}{n ( n + 3 )}

\frac{11}{3}

Solution steps

n = 1 \sum \infty \frac{6}{n ( n + 3 )}

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 + 3 )}

Take the partial fraction of

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

= 6 \cdot n = 1 \sum \infty \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

Expand telescoping series:

\frac{11}{18}

Simplify

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

= \frac{11}{3}

\frac{d}{d x} (2 x + cos (2 x))

\int sin (\frac{nπ x}{L}) d x

f (t) = (t + 1)^{\frac{2}{3}} (2 t^{2} - 1)^{3}

\int \frac{1}{sec ( x ) tan ( x )} d x

\frac{d}{d t} (\frac{2 t + 5}{2 t})

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