Question 1: Partial Fractions

$\begin{aligned} S & = {\color{#3D99F6}\frac 12} + {\color{#D61F06}\frac 16} + {\color{#3D99F6}\frac 1{12}} + {\color{#D61F06}\frac 1{20}} + \cdots \\ & = {\color{#3D99F6} \frac 1{1\cdot 2}} + {\color{#D61F06} \frac 1{2\cdot 3}} + {\color{#3D99F6} \frac 1{3\cdot 4}} + {\color{#D61F06} \frac 1{4\cdot 5}} + \cdots \\ & = {\color{#3D99F6} \frac 11 - \frac 12} + {\color{#D61F06} \frac 12 - \frac 13} + {\color{#3D99F6} \frac 13 - \frac 14} + {\color{#D61F06} \frac 14 - \frac 15} + \cdots \\ & = {\color{#3D99F6} \frac 11 -\cancel{\frac 12}} + {\color{#D61F06}\cancel{\frac 12} - \cancel{\frac 13}} + {\color{#3D99F6} \cancel{\frac 13} - \cancel{\frac 14}} + {\color{#D61F06} \cancel{\frac 14} - \cancel{\frac 15}} + \cdots \\ & = \boxed{1} \end{aligned}$

It is always $1$ because $\displaystyle \frac { 1 } { 2 } + \frac { 1 } { 6 } + \frac { 1 } { 12 } + \frac { 1 } { 20 } + \cdots = 1$ .

That is because all fractions are less than one, and the numerators and denominators are all positive.

$\small{ S = \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + \cdots = \dfrac{1}{2}\left( \dfrac{1}{1}+ \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{10} + \cdots \right) \\ S = \dfrac{1}{2}\left( \dfrac{1}{1}+ \dfrac{1}{1+2} + \dfrac{1}{1+2+3} + \dfrac{1}{1+2+3+4} + \cdots \right) \\ S = \dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{2}{n(n+1)} = \dfrac{2}{2} \underbrace{\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n+1}\right)}_{\text{Telescoping sum}}= \boxed{1}}$