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

`\frac1k-\frac1{k+1}=(k+1)/(k(k+1))-\frac k{k(k+1)}=(k+1-k)/(k(k+1))=\frac1{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-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1{2015}-\frac1{2016}\right)+\left(\frac1{2016}-\frac1{2017}\right)`

All the intermediate terms cancel, leaving you with


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