Sum of this infinite series

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

Each ot the terms in the expression can be generalized as Sum to infinite terms of 1/n*(n+1) where n ranges from 1 to infinity.

1/n*n+1 = 1/n - 1/n+1.

So the expression the question can be rewritten as

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5......-1/k + 1/k - 1/k+1

= 1 - 1/k + 1

When k tends to infinity this reduces to 1.

So the sum of all these terms in the question is equal to 1.