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$\begin{aligned} \sum_\blue{n=2}^{2021} \frac 1{n(n-1)} & = \sum_\red{n=1}^{2020} \frac 1{n(n+1)} & \small \blue{\text{By partial fraction decomposition}} \\ & = \sum_{n=1}^{2020} \left(\frac 1n - \frac 1{n+1} \right) \\ & = \frac 11 \ \cancel{- \frac 12 + \frac 12} \ \cancel{- \frac 13 + \frac 13} \ \cancel{- \frac 14 + \frac 14} - \cdots \cancel{- \frac 1{2020} + \frac 1{2020}} - \frac 1{2021} \\ & = 1 - \frac 1{2021} = \boxed {\frac {2020}{2021}} \end{aligned}$

$\sum_{n=2}^{2021} \dfrac {1}{ (n-1)(n)} = 1 - \frac { 1 }{2021} = \frac{2020}{2021}$

We can write the first few terms to see how terms get cancelled:

$\frac{1}{1\times 2} = 1 - \frac{1}{2}$

$\frac{1}{2\times 3} = \frac{1}{6} = \frac{1}{2} - \frac{1}{3}$

$\frac {1}{3\times 4} = \frac {1}{12} = \frac {1}{3} - \frac{1}{4}$

$etc ...$

$\sum_{n=2}^{2021} \dfrac {1}{ (n-1)(n)} =\Big ( 1 - \cancel{\frac{1}{2}}\Big) +\Big ( \cancel{\frac{1}{2}} -\cancel{ \frac{1}{3}}\Big) + \Big( \cancel{\frac {1}{3}} -\cancel{ \frac{1}{4}}\Big)+\dots +\Big(\cancel{\frac{1}{2020}} - \frac{1}{2021} \Big) = 1 - \frac{1}{2021} = \frac{2020}{2021}$