What is the sum from n=0 to infinity of 1/(n(n+3)) ?

The sum from n=0 to infinity of 1/(n(n+3)) is 11/18

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sum from n=0 to infinity of 1/(n(n+3))

n = 0 \sum \infty \frac{1}{n ( n + 3 )}

\frac{11}{18}

Solution steps

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 )}

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

Expand telescoping series:

\frac{11}{18}

= \frac{11}{18}

\int \frac{x ^{3}}{x ^{2} - 25} d x

\frac{\partial}{\partial x} (x^{2} - 2 x + 3)

\frac{d}{d x} (\frac{4 x}{3})

\frac{8}{1 + x ^{4}}, 4 x^{2}

y^{^{'}} + \frac{y}{t} = 3 cos (2 t)

What is the sum from n=0 to infinity of 1/(n(n+3)) ?
The sum from n=0 to infinity of 1/(n(n+3)) is 11/18