What is the sum from n=1 to infinity of 1/((n(n+2))) ?

The sum from n=1 to infinity of 1/((n(n+2))) is 3/4

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sum from n=1 to infinity of 1/((n(n+2)))

n = 1 \sum \infty \frac{1}{( n ( n + 2 ) )}

\frac{3}{4}

Solution steps

n = 1 \sum \infty \frac{1}{n ( n + 2 )}

Take the partial fraction of

\frac{1}{n ( n + 2 )} : \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

= n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

= \frac{3}{4}

\int (\frac{1}{x ^{- 0.1}}) d x

y^{^{'}} = (x + y - 1)^{2}

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

(arctan (9 x))^{2}

\frac{e ^{x}}{5 x - 2}

What is the sum from n=1 to infinity of 1/((n(n+2))) ?
The sum from n=1 to infinity of 1/((n(n+2))) is 3/4