Question

Simplify 1/k-1/(k+1). Hence or otherwise, use the above to find a value for ∑_(k=1)^2017▒1/(k(k+1))

Answer 1

$\dfrac1k-\dfrac1{k+1}=\dfrac{k+1}{k(k+1)}-\dfrac k{k(k+1)}=\dfrac{k+1-k}{k(k+1)}=\dfrac1{k(k+1)}$

This means we can write the given sum as a telescoping series:

$\displaystyle\sum_{k=1}^{2017}\frac1{k(k+1)}=\sum_{k=1}^{2017}\left(\frac1k-\frac1{k+1}\right)$
$=\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\cdots+\left(\dfrac1{2015}-\dfrac1{2016}\right)+\left(\dfrac1{2016}-\dfrac1{2017}\right)$

All the intermediate terms cancel, leaving you with

$\displaystyle\sum_{k=1}^{2017}\frac1{k(k+1)}=1-\frac1{2017}=\frac{2017}{2018}$