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}

y = x^{5}

x \to 3 lim (\frac{x + 2}{x - 2})

\frac{d}{d x} (cos (\frac{π}{3}))

x^{2} + 10 x + c

\frac{1}{s ^{3} ( s + 3 )}

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