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

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

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

n = 1 \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 1 - 2 x d x

a^{2} + 2 a + 2

\frac{d}{d x} (sech (2 x))

\frac{\partial}{\partial x} ((3 x + 2 y)^{5})

x \to 4 lim (\frac{1}{x - 4})

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