Spot the pattern

The $n$ th term of the series is $a_{n} = \dfrac{1}{n(n + 2)} = \dfrac{1}{2}\left(\dfrac{1}{n} - \dfrac{1}{n + 2}\right).$

Thus $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n(n + 2)} = \dfrac{1}{2} \sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n + 2}\right) = \dfrac{1}{2}\left(1 + \dfrac{1}{2} + \sum_{n=3}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n}\right)\right) = \dfrac{1}{2} * \dfrac{3}{2} = \boxed{0.75}.$

We have a general pattern: $\dfrac{1}{n (n+2)} \\ =\dfrac {1}{2}\left (\dfrac{1}{n}-\dfrac {1}{n+2}\right)$ Our expression becomes: $\dfrac {1}{2}\left (\dfrac {1}{1}-\dfrac {1}{3}\right)+\dfrac {1}{2}\left (\dfrac {1}{2}-\dfrac {1}{4}\right)+\dfrac {1}{2}\left (\dfrac {1}{3}-\dfrac {1}{5}\right)\ldots \\ =\dfrac {1}{2}\left (1-\dfrac {1}{3}+\dfrac {1}{2}-\dfrac {1}{4}+\dfrac {1}{3}-\dfrac {1}{5}\ldots\right) \\ =\dfrac {1}{2}\left (1+\dfrac {1}{2}\right) \\ =\boxed {\dfrac {3}{4}}$

The following sum can be expressed as $\displaystyle\sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r\left( r+2 \right) } }$ . We will try to split this into two separate fractions:

Suppose $\dfrac { 1 }{ r\left( r+2 \right) } =\dfrac { A }{ r } +\dfrac { B }{ r+2 } \Rightarrow 1=A\left( r+2 \right) +Br$ .

Let $r=-2$ . Then $1=-2B \Rightarrow B=-\dfrac { 1 }{ 2 }$ .

Let $r=0$ . Then $1=2A \Rightarrow A=\dfrac { 1 }{ 2 }$ .

$\therefore ~~~ \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r\left( r+2 \right) } } = \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ 2r } -\frac { 1 }{ 2\left( r+2 \right) } } \\ = \left( \frac { 1 }{ 2 } -\frac { 1 }{ 6 } \right) +\left( \frac { 1 }{ 4 } -\frac { 1 }{ 8 } \right) +\left( \frac { 1 }{ 6 } -\frac { 1 }{ 10 } \right) +\left(\ \frac { 1 }{ 8 } -\frac { 1 }{ 12 } \right) +\dots \\ = \frac { 1 }{ 2 } +\frac { 1 }{ 4 } \\ = ~~ \large \color{#20A900}{\boxed{\frac{3}{4}}}$